8.1: Ear Training
- Ear training develops the basic listening skills of the music student. Here are some suggestions for developing the various ear training skills, including a downloadable game to play.
What is Ear Training?
When musicians talk about ear, they don't mean the sense organ itself so much as the brain's ability to perceive, distinguish, and understand what the ear has heard. The term ear training refers to teaching musicians to recognize information about notes and chords just by hearing them.
A few people have what is called perfect pitch or absolute pitch. These people, when they hear music, can tell you exactly what they are hearing: the G above middle C, for example, or the first inversion of an F minor chord. A few musicians with particularly perceptive ears can even tell you that a piano is tuned a few cents higher than the one that they play at home. This is an unusual skill that even most trained musicians do not have, and research seems to suggest that if you don't have it at a very early age, you cannot develop it. (For more on this subject, you may want to look up Robert Jourdain's Music, the Brain, and Ecstasy: How Music Captures our Imagination .)
However, most musicians can be trained to recognize relative pitch. In other words, if you play two notes, they can tell you that one of them is a major third higher than the other. If you play four chords in a row, they can tell you that you played a tonic-subdominant-dominant seventh-tonic (I-IV-V7-I) chord progression.
Fortunately, having relative pitch is good enough, and for many musicians may even be more useful than perfect pitch, because of the way Western music is conceived. Since all major keys are so similar, a piece in a major key will sound almost exactly the same whether you play it in C major or D major. The thing that matters is not what note you start on, but how all the notes are related to each other and to the "home" note (the tonic) of the key. If someone really wants the piece to be in a different key (because it's easier to sing or play in that key, or just because they want it to sound higher or lower), the whole thing can be transposed, but the only difference that would make (in the sound) is that the entire piece will sound higher or lower. Most listeners would not even notice the difference, unless you played it in both keys, one right after the other.
All minor keys are also heard by most listeners as interchangeable, but there are important differences between major keys and minor keys. In fact, the differences in sound between a major key and a minor key is one of the first differences that a musician should be able to hear. If you would like to see whether your "ear" can recognize the difference between major and minor keys, please try the listening exercise in Major Keys and Scales.
So, you often don't need to know exactly what notes or chords are being played. Simply having an ear well-trained in "relative pitch" is extremely useful in many ways. Guitar and piano players can figure out chord progressions just by listening to them, and then play the progressions in their favorite keys. Other instrumentalists can play a favorite tune without a written copy of it, just by knowing what the interval to the next note must be. Composers and music arrangers can jot down a piece of music without having to "pick it out" on an instrument to find the notes and chords they want. And of course, ear training is crucial to any musician who wants to play jazz or any type of improvisation. Given a well-trained "ear", any musical idea that you "hear" in your head, you can play. And ear training is also crucial for those interested in music theory, musicology, or just being able to write down a tune accurately.
As with all other musical skills, there are many different levels and kinds of proficiency. One musician may be very good at "playing by ear", but may not even read music and cannot name intervals or write the music down. Another may be very good at "taking dictation" (writing down the music they hear), and yet feel unable to do jazz improvisation. As always, the key is to practice the particular skills that you want to develop.
Ear Training Skills
Tuning
This is the most basic ear training skill, crucial to being able to play music that people will want to hear.
Suggestions
- At the beginner level, work with a skilled musician who can teach you how to tune your instrument and help you identify and fix tuning problems.
- Play with other musicians often. (Playing along with recordings does not teach good tuning skills.) Don't just tune at the beginning of rehearsals and performances. Listen at all times and be ready to retune any note whenever necessary.
- Spend as much time as necessary tuning whenever you play. Do not (knowingly) practice while out of tune; if you do, it will slow down your ear training tremendously. Whenever possible, until you are good at tuning, get someone else to help you tune every time you play.
- Practice tuning quickly and accurately. Learn any alternate fingerings and other "tricks" available on your instrument for fine-tuning each note as you play.
Playing Chords By Ear
For instruments that play chordal accompaniments, this is an incredibly useful skill.
Suggestions
- You do not have to learn to read music to be able to do this, but it is very helpful to know a little bit about music theory so that you can predict which chords are most likely to happen in a song. Try starting with Beginning Harmonic Analysis.
- Really listen to the chord progressions to the songs you do know. What do they sound like? Play the same progressions in different keys and listen to how that does and also does not change the sound of the progression. Change the bass notes of the chords to see how that changes the sound of the progression to your ears. Change fingerings and chord voicings, and again listen carefully to how that changes the sound to your ears.
- Practice figuring out the chords to familiar songs (that you don't know the chords to). For songs that you do know the chords to, try playing them in an unfamiliar key, or see if you can change or add chords to make a new harmony that still fits the melody.
- A teacher who understands harmony can help tremendously with this particular skill. Even if you don't normally take lessons, you might want to consider having a series of lessons on this. Find a teacher who is willing and able to teach you specifically about harmony and typical chord progressions.
Playing Tunes by Ear
This is fun to be able to do, makes it easy to increase your repertoire, and is an important step in being able to improvise.
Suggestions
- Just do it! The best way to learn this skill is to spend some of your practice time trying to play tunes you know and like.
- Once you start getting good at this, see how quickly you can get a new tune down. How few mistakes can you make the first time you try it? Can you "recover" quickly from a mistake by making it sound like a bit of improvisation?
- If you play a melody instrument (one that plays only one note at a time), there are different bits of information that help you recognize what the next note will be: how far it is from the note you are on (see Interval), where it is in the key (see Beginning Harmonic Analysis) or where it is in the chord (see Triads). These three things are all related to each other, of course - and a musician with a well-trained ear will be aware of all of them, at least subconsciously - but you may find at first that one works better for you than the others. You may want to experiment: is it easier for you to think of the next note as being a perfect fourth higher than the note you are on, or as being the root of the chord, or as being the fifth note in the scale of the key?
- As of this writing, petersax-online had many exercises graded from simple to more difficult to help the beginner practice playing what you hear.
Improvisation
This is the skill you need for jazz. Blues, rock, and many Non-Western traditions also use improvisation.
Suggestions
- Know your scales and arpeggios. A good improviser, given the name of a chord, can quickly play not only the notes of the chord but also the scale implied by the chord. Any decent book on playing jazz, or any teacher familiar with jazz, will introduce the student to these chords and scales.
- There are now many book/CD combinations available to help the beginning improviser in many different genres and on many different instruments. A good book of this type will give the student a chance to improvise on many familiar tunes, and some also introduce the music theory involved. At the time of this writing, one source of a large variety of such books was jazzbooks.com .
- The exercises at the petersax site mentioned above would also be useful for the beginning improviser.
- Listen to jazz often. Listen to the improvisers you admire, and if a particular solo really appeals to you, listen to it many times, find the notes on your instrument, and then try writing it down as accurately as you can. Many famous improvisors, when interviewed, mention how useful it was to them to learn from other soloists by transcribing their solos in this way.
- Figure out how to play your favorite jazz (or blues or rock) licks (short motives that show up in many pieces in the same genre) on your instrument. Practice stringing them together in ways that make sense to you, but are different from what you've heard. Add your own variations.
- Find a teacher who is familiar with the type of improvisation you want to learn, join a jazz band, and/or get together with other musicians who also want to practise improvisation and take turns playing background/rhythm for each other.
Recognizing Intervals and Writing Music Down
This is the skill that allowed Beethoven to continue composing masterpieces even after he became deaf. If you are interested in composing, arranging, music theory, musicology, or just being able to write down a tune quickly and accurately, you'll want to be able to make that quick connection between what you hear and written music.
Suggestions
- Before you can do this, you must know your major and minor keys and scales and your Intervals. You may also want to understand Transposition, since you may find it easier to work in some keys than in others.
- As of this writing, Teoria Musical was a free ear training website that worked well, and the commercial site TrainEar included a free online version.
- Once again, practice is the best way to become good at this. Start with tunes that you know well, but don't know what the (written) notes are. Listen to them in your head (or play a recording) while trying to write them down. Then play what you have written, noticing where you were correct and where you made mistakes. Which intervals are you good at hearing? Which do you have trouble identifying? Do you often mistake one particular interval for another? Do you tend to identify a note by its interval from the previous note or by its place in the chord or in the key? Answering these questions will help you improve more quickly.
- Some people find it easier to learn to recognize intervals if they associate each interval with a familiar tune. (For example, in the familiar song from The Sound of Music that begins "Do, a deer, a female deer...", all the intervals in the phrase "a female deer" are major thirds, and every interval in the phrase "someday I'll wish upon a star" in the song "Somewhere Over the Rainbow" is a minor third.) The tune should be very familiar, so when trying to hear a tritone, some people will prefer thinking of the beginning of "The Simpsons" theme; others will prefer the beginning of "Maria" from West Side Story . If you think this method will work for you, try playing the interval you are having trouble hearing, and see what tune it reminds you of. As of this writing, TrainEar included a long list, with links to recordings, or songs that can be associated with various intervals.
- Try searching at YouTube for "Interval song" or "ear training" to find videos that you might find helpful.


29.4C: Evolution of Reptiles
- Outline the evolution of reptiles
Evolution of Reptiles
Reptiles originated approximately 300 million years ago during the Carboniferous period. One of the oldest-known amniotes is Casineria , which had both amphibian and reptilian characteristics. One of the earliest undisputed reptiles was Hylonomus . Soon after the first amniotes appeared, they diverged into three groups (synapsids, anapsids, and diapsids) during the Permian period. The Permian period also saw a second major divergence of diapsid reptiles into archosaurs (predecessors of crocodilians and dinosaurs) and lepidosaurs (predecessors of snakes and lizards). These groups remained inconspicuous until the Triassic period when the archosaurs became the dominant terrestrial group due to the extinction of large-bodied anapsids and synapsids during the Permian-Triassic extinction. About 250 million years ago, archosaurs radiated into the dinosaurs and the pterosaurs.
Although they are sometimes mistakenly called dinosaurs, the pterosaurs were distinct from true dinosaurs. Pterosaurs had a number of adaptations that allowed for flight, including hollow bones (birds also exhibit hollow bones, a case of convergent evolution). Their wings were formed by membranes of skin that attached to the long, fourth finger of each arm and extended along the body to the legs.
The dinosaurs were a diverse group of terrestrial reptiles with more than 1,000 species identified to date. Paleontologists continue to discover new species of dinosaurs. Some dinosaurs were quadrupeds; others were bipeds. Some were carnivorous, whereas others were herbivorous. Dinosaurs laid eggs; a number of nests containing fossilized eggs have been found. It is not known whether dinosaurs were endotherms or ectotherms. However, given that modern birds are endothermic, the dinosaurs that served as ancestors to birds were probably endothermic as well. Some fossil evidence exists for dinosaurian parental care. Comparative biology supports this hypothesis since the archosaur birds and crocodilians display parental care.
Dinosaurs dominated the Mesozoic Era, which was known as the “Age of Reptiles.” The dominance of dinosaurs lasted until the end of the Cretaceous period, the end of the Mesozoic Era. The Cretaceous-Tertiary extinction resulted in the loss of most of the large-bodied animals of the Mesozoic Era. Birds are the only living descendants of one of the major clades of dinosaurs.
Key Points
- Diapsids diverged into archosaurs and lepidosaurs, but these groups did not dominate the ecosystem until the Triassic following the Permian extinction.
- Archosaurs diverged into the dinosaurs and the pterosaurs about 250 million years ago.
- Pterosaurs had the ability to fly because of their wings and hollow bones, a trait convergent to modern birds, but were not ancestral to birds.
- Dinosaurs were quadrupeds or bipeds, carnivorous or herbivorous, and laid eggs.
- It is unknown whether dinosaurs were endothermic or ectothermic, but since birds are endothermic, the dinosaur ancestors of birds were probably endothermic.
- Dinosaurs dominated the Mesozoic Era until the Cretaceous -Tertiary extinction wiped out most of these large-bodied animals.
Key Terms
- pterosaur : any of several extinct flying reptiles, of the order Pterosauria, including the pterodactyls
- Cretaceous-Tertiary extinction : mass extinction of three-quarters of plant and animal species on earth, including all non-avian dinosaurs, that occurred over a geologically-short period of time 66 million years ago


<IP_ADDRESS>: Commerce Clause
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The Constitution and the Law
Federal and state constitutions are a major source of business law. The United States Constitution is the supreme law of the United States. In addition to the individual constitutions established in each state, the U.S. Constitution sets out the fundamental rules and principles by which the country and individual states are governed. Constitutional law is the term used to describe the powers and limits of the federal and state governments as established in the Constitution. The political system that divides authority to govern between the state and federal governments is known as federalism , and this too is established in the Constitution. The Tenth Amendment states that any area over which the federal government is not granted authority through the Constitution is reserved for the state. This statement means that any federal legislation impacting business and commerce must be established by an expressed constitutional grant of authority .
Federal Preemption
The Founding Fathers created a federal system that would, at times, “preempt” state law through the supremacy clause , outlined in Article VI of the Constitution. In other words, since the U.S. Constitution is the “supreme law of the land,” if a state law conflicts with the U.S. Constitution, the state law is declared invalid. When the federal constitutional law prevails over the state law, it is said that the state law has been preempted . Before that determination is made, the courts try to determine if Congress intended to preempt state law in enacting the particular provision in question. If the answer is “no,” then those who are asserting protections of state law may make claims under state law. If the answer is “yes,” however, federal law prevails.
The Tenth Amendment to the Constitution gives the states powers over areas of law not held exclusively by the federal government through the U.S. Constitution, e.g., states can make laws about how to get married, who may get married, or how to dissolve a marriage, as well as which activities are crimes and how the crimes will be punished. If the U.S. Constitution does give the federal government some power, however, then the federal government may exercise it, free from state interference. For instance, the U.S. Congress (the legislative branch of the federal government) has the power, among other things, to coin money, to create a military, to establish post offices, and to declare war. Since there is specific mention of these powers, states may not create their own currency, military, or postal service, and they may not declare war.
The Commerce Clause and The Affordable Care Act
After much debate, negotiation, and political wrangling, Congress passed the Patient Protection and Affordable Care Act (PPACA) in 2010, which was designed to increase the number of Americans who had access to health insurance (a policy initiative known as Obamacare). The Act included a provision mandating that individuals not insured through employment or who were otherwise exempt from receiving health insurance obtain minimum essential health insurance or face a penalty issued through the Internal Revenue Service (IRS). The National Federation of Independent Business (NFIB), supported by \(26\) of the \(50\) states, challenged the constitutionality of this particular provision, known as the individual mandate. Their argument was upheld by the 11th Circuit Court of Appeals, which ruled that Congress did not have the authority to enact this provision. Later, however, the appellate court determined that the individual mandate was severable from the remainder of the PPACA, so ultimately the Act was upheld.
The main source of authority for the federal regulation of interstate and international commerce is the commerce clause . This clause is established in Article I, Section 8, of the Constitution. The Article grants Congress the power to “regulate Commerce with foreign Nations, and among the several States, and with the Indian Tribes.” Thus, the commerce clause serves to simultaneously empower the federal government, while limiting state power.
So long as a federal regulation impacts interstate commerce, that regulation can be described as constitutional, according to the commerce clause. However, since the Constitution was first written, there have often been occasions when the judiciary system has needed to step in to interpret the meaning and implications of the commerce clause. In particular, there have been disputes over the intended meaning of the phrase “among the several States.” Up until the 1930s, this phrase was interpreted in a literal way, so that activities subject to federal regulation were required to involve trade between the states. This strict interpretation actually served to limit the federal regulation of commerce.
The turning point in the interpretation of the commerce clause came with the 1937 case, NLRB v. Jones & Laughlin Steel Corp . The previous year, in the Carter v. Carter Coal Co case, the court invalidated a program, initiated under the New Deal, that had tried to regulate the labor practices of coal firms on the basis that these practices were local, and therefore had only an indirect impact on interstate commerce. In NLRB v. Jones & Laughlin Steel Corp , the court deviated from that decision by ruling that Congress could regulate employment practices at a steel plant because any stoppages at that plant would have a serious, detrimental impact on interstate commerce. The court concluded that since the steel industry is a networked industry that incorporates mines, plants, and factories from Minnesota to Pennsylvania, the manufacturing of steel properly falls under the jurisdiction of the commerce clause. In summing up, the court concluded that:
“Although activities may be intrastate in character when separately considered, if they have such a close and substantial relationship to interstate commerce that their control is essential or appropriate to protect that commerce from burdens or obstructions, Congress cannot be denied the power to exercise that control” (NLRB v. Jones & Laughlin Steel Corp., 301 U.S. 1 1937).
Challenges to and Reinterpretations of the Commerce Clause
Ever since the NLRB v. Jones & Laughlin Steel Corp case, Congress has invoked the commerce clause to rule on a diverse range of business and commercial activities, as well as to support social reforms that indirectly impact state commerce. Examination of the United States Code reveals that there are more than \(700\) legislative provisions that explicitly refer to foreign or interstate commerce. What is perhaps most remarkable is the sheer diversity of statutory areas covered by the commerce clause. Areas covered include the regulation of sporting activities, endangered species, energy regulation, gambling, firearms control, and even terrorism.
Examples of Federal Legislation Passed by Invoking the Commerce Clause
- The Controlled Substances Act
- The Federal Mine Safety and Health Act
- The Civil Rights Act
- Americans with Disabilities Act
- The Indian Child Welfare Act
While businesses have often challenged these statutes as existing outside of the realm of congressional authority, in most cases, the courts have upheld the statutes as being valid exercises of congressional power in line with the commerce clause. An exception is the 1995 case, United States v. Lopez . The case centered around the legality of the Gun-Free School Zone Act, which was a federal law that outlawed the possession of guns within 1,000 feet of a school. In a landmark case, the Court ruled that the Act was outside the scope of the commerce clause, and that Congress did not have the authority to regulate in an area that had “nothing to do with commerce, or any sort of enterprise.”
A recent controversy pertaining to the commerce clause relates to the passing of the Affordable Care Act, as described earlier. Protestors claimed that the individual mandate aspect of the ACA should be treated as a regulation that affects interstate commerce. According to their argument, after the Act was implemented, there would be an increase in the sale and purchase of health care insurance, such that the market for health care should be seen as being significantly impacted by the Act. However, the Chief Justice of the Supreme Court, Justice Roberts, ruled that actions that create new business activity do not affect interstate commerce.
Police Power and the Dormant Commerce Clause
The authority of the federal government to regulate interstate commerce has, at times, come into conflict with state authority over the same area of regulation. The courts have tried to resolve these conflicts with reference to the police power of the states.
Police power refers to the residual powers granted to each state to safeguard the welfare of their inhabitants. Examples of areas in which states tend to exercise their police power are zoning regulations, building codes, and sanitation standards for eating places. However, there are times when the states’ use of police power impacts interstate commerce. If the exercise of the power interferes with, or discriminates against, interstate commerce, then the action is generally deemed to be unconstitutional. The limitation on the authority of states to regulate in areas that impact interstate commerce is known as the dormant commerce clause .
In using the dormant commerce clause to resolve conflicts between state and federal authority, the courts consider the extent to which the state law has a legitimate purpose. If it is determined that the state law has a legitimate purpose, then the court tries to determine whether the impact on interstate commerce is in the interest of the citizens of the state, and will rule accordingly. For instance, an ordinance that banned spray paint, issued in the city of Chicago, was challenged by paint manufacturers under the dormant commerce clause, but was ultimately upheld by the U.S. Court of Appeals because the ban was intended to reduce graffiti and related crimes.
Today, Congress uses its authority to regulate commercial activity in four general areas relating to the commerce clause:
- Regulation of the channels of interstate commerce
- Regulation of the instrumentalities of interstate commerce
- Regulation of intangibles and tangibles that cross state lines
- Regulation of activities that are deemed to be both economic and to have a substantial impact on interstate commerce
| Area of Regulation | Explanation | Examples |
|---|---|---|
| Regulation of the channels of interstate commerce | Channels of interstate commerce describe the passages of transportation between the states. Thus, the commerce clause authorizes Congress to regulate activities pertaining to the nation’s airways, waterways, and roadways, and even where the activity itself takes place entirely in a single state. | For example, Congress can pass regulations that restrict what can be carried on airlines or on ships. |
| Regulation of the instrumentalities of interstate commerce | Instrumentalities of commerce are understood to be any resource employed in the carrying out of commerce. Examples of these resources are machines, equipment, vehicles, and personnel. Thus, Congress has the power to regulate these areas. | Congress could pass regulations mandating certain safety standards for equipment used in manufacturing plants. |
| Regulation of intangibles and tangibles that cross state lines | Any object, tangible or intangible, that crosses state lines can be regulated under the commerce clause. Tangible objects include goods purchased by consumers, as well as raw materials and equipment used in the production of goods for sale. Intangible objects include services, as well as electronic databases. | The Driver’s Privacy Protection Act (DPPA) regulates the sale of information contained in the Department of Motor Vehicles’ (DMV’s) records. |
| Regulation of activities that are deemed to have a substantial impact on interstate commerce | Federal regulation of economic commercial activity expected to have a significant (as opposed to minor) effect on interstate commerce is constitutional, according to the commerce clause. Noneconomic commercial activity is not covered. | The courts in the United States vs. Lopez case described earlier deemed the Act to be unconstitutional because its terms have “nothing to do with ‘commerce’ or any sort of economic enterprise.” |
Contributors and Attributions
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The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the creative commons license and may not be reproduced without the prior and express written consent of Rice University. For questions regarding this license, please contact<EMAIL_ADDRESS>. Download for free at https://openstax.org/details/books/b...w-i-essentials .


20.1: Introduction
Musculoskeletal and integumentary changes can affect other organ systems, and they also become evident as the body ages. Aging muscles and skin can cause changes in a person’s mobility and overall wellness. Decreases in bone mass and density can result in fractures and illness. The chapter on Applying Clinical Judgment to Promote Nutrition for Musculoskeletal and Integumentary Wellness covers the important role of nutrition in musculoskeletal and integumentary health and wellness. This chapter will introduce special considerations across the lifespan because nutritional needs change over time. Preventing and mitigating complications of the musculoskeletal and integumentary systems, such as fractures, poor wound healing, physical disability, disease progression, poor quality of life, and hospital admissions, can often be achieved through implementing appropriate nutrition and adequate physical activity.
Consider this case: Meet Maria, a 20-year-old, and her newborn baby, Jose. To explore the concepts presented in this chapter, the unfolding case study will follow Jose as he grows from a newborn to an older adult.


6.17: Getting Started
- Explain strategies to help begin your essay
Drafting
At this point in the writing process, you’ll have accomplished quite a lot of work. It may feel pretty scattered across notes you’ve made, sources you’ve pulled together, and different trains of thought in your head. That’s fine!
You’ll likely have some sort of outline at this point: some plan for what sections you’ll need, and an idea of their order. This may be a very formal Roman numeral outline, a more informal list of ideas, a mind map, a PowerPoint outline—whatever is comfortable for you.
The next step is a big one: starting to flesh out those pieces of an outline into a substantial essay draft. Drafting includes prewriting, editing, and reviewing. Once your general ideas are down on paper, writing out specific ideas and quotations can make the final writing process much easier. Each step of drafting brings the process a little closer to the final product.
Always write down any ideas you have in the drafting process—it is much easier to cut content from your paper than it is to work on adding content. If you collect all your resources, quotations, facts, ideas, and come up with a thesis during the drafting process, your paper will show it. The idea is to provide yourself with as much information as possible in order to create a solid and well-thought-out piece. Do less worrying and more writing. Get started as soon as you possibly can! The first draft can be ugly and messy—the key is to get something out on paper (or screen) that you can enhance later on.
This video explains some techniques for getting starting on your first draft. You may want to start by writing a “fast draft,” which involves writing quickly, or “blurting” to get your thoughts out on the page. When you write a fast draft, you can use brackets to leave notes to yourself about areas where you’ll want to come back and add more detail or find research, etc.
Slow drafts, in which you worry more about details and sentence structure, are more time consuming but may work well if you prefer to dive in deep from the beginning. Even with a slow draft, you should still expect to work through many revisions of the paper.
You can view the transcript for “How to Plan and Write a Paper: How to Draft” here (opens in new window) .
They Say, I Say
Sometimes looking at the template and organization of other types of papers can help you get started on your own. Gerald Graff and Cathy Birkenstein published a helpful book, They Say/I Say [1] that provides templates of actual language you could use in written arguments. The title of the book comes from the idea that writing is a form of joining an existing conversation—you find out what others are talking about, you add in your own two cents, and the conversation continues in various forms even after you leave. When writing, you are entering into a conversation with other scholars and other authors who have considered your same topic, so you can approach your paper by first considering what, “They say,” and then adding in your own, “I say.” For example, in starting your argument, you could say something like:
- X argues that…
- X claims that…
- X demonstrates that…
Then you could support or contradict those ideas with your own beliefs. Templates for joining in an ongoing discussion may look something like this:
- In discussions of __________, one controversial issue has been ____________. On the one hand, ______________ argues __________________. On the other hand, ______________ opposes _____________. Others even believe ______________. My own view is ______________.
- When it comes to the topic of ________, most of us will readily agree that __________. Where this agreement usually ends, however, is on the question of _____________. While some are convinced that _____________, others believe that _____________. [2]
You can help readers follow your argument by inserting phrases that signal why you’re doing what you’re doing. Here are some examples:
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To state that what you’re saying in your thesis (answer to your research question) is in opposition to what others have said:
- “Many people have believed …, but I have a different opinion.”
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To move from a reason to a summary of a research study that supports it (evidence):
- “Now let’s take a look at the supporting research.”
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To introduce a summary of a resource you’ve just mentioned:
- “The point they make is…”
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To acknowledge an objection you believe a reader could have:
- “At this point I should turn to an objection some are likely to be raising…”
Phrases like these can grease the skids of your argument in your readers’ minds, making it a lot easier for them to quickly get it instead of getting stuck on figuring out why you’re bringing something up at a particular point. You will have pulled them into an argument conversation.
Using templates such as this takes some of the mystery out of what to say and when to say it. To see more of these templates, purchase the book, or check it out from your library.
The blog that accompanies the book They Say/I Say with Readings, by Gerald Graff, Cathy Birkenstein, and Russel Durst, contains short, elegantly constructed contemporary arguments from a variety of publications. Take a look at the They Say/I Say blog for a moment and read part of at least one of the readings to see how it can be helpful to you the next time you have to make a written argument.
Contributors and Attributions
- Helping Others Follow. Provided by : Ohio State University Libraries. Located at : https://ohiostate.pressbooks.pub/choosingsources/chapter/helping-others-follow/ . Project : Choosing & Using Sources: A Guide to Academic Research . License : CC BY: Attribution
- Modification, adaptation, and original content. Provided by : Lumen Learning. License : CC BY: Attribution


5.9: Exercises
Explain how σ and π bonds are similar and how they are different.
Use valence bond theory to explain the bonding in F 2 , HF, and ClBr. Sketch the overlap of the atomic orbitals involved in the bonds.
Use valence bond theory to explain the bonding in O 2 . Sketch the overlap of the atomic orbitals involved in the bonds in O 2 .
How many σ and π bonds are present in the molecule HCN?
A friend tells you N 2 has three π bonds due to overlap of the three p -orbitals on each N atom. Do you agree?
Draw the Lewis structures for CO 2 and CO, and predict the number of σ and π bonds for each molecule.
(a) CO 2
(b) CO
Why is the concept of hybridization required in valence bond theory?
Give the shape that describes each hybrid orbital set:
(a) sp 2
(b) sp 3 d
(c) sp
(d) sp 3 d 2
Explain why a carbon atom cannot form five bonds using sp 3 d hybrid orbitals.
What is the hybridization of the central atom in each of the following?
(a) BeH 2
(b) SF 6
(c)
(d) PCl 5
A molecule with the formula AB 3 could have one of four different shapes. Give the shape and the hybridization of the central A atom for each.
Methionine, CH 3 SCH 2 CH 2 CH(NH 2 )CO 2 H, is an amino acid found in proteins. The Lewis structure of this compound is shown below. What is the hybridization type of each carbon, oxygen, the nitrogen, and the sulfur?
Sulfuric acid is manufactured by a series of reactions represented by the following equations:
Draw a Lewis structure, predict the molecular geometry by VSEPR, and determine the hybridization of sulfur for the following:
(a) circular S 8 molecule
(b) SO 2 molecule
(c) SO 3 molecule
(d) H 2 SO 4 molecule (the hydrogen atoms are bonded to oxygen atoms)
Two important industrial chemicals, ethene, C 2 H 4 , and propene, C 3 H 6 , are produced by the steam (or thermal) cracking process:
For each of the four carbon compounds, do the following:
(a) Draw a Lewis structure.
(b) Predict the geometry about the carbon atom.
(c) Determine the hybridization of each type of carbon atom.
Analysis of a compound indicates that it contains 77.55% Xe and 22.45% F by mass.
(a) What is the empirical formula for this compound? (Assume this is also the molecular formula in responding to the remaining parts of this exercise) .
(b) Write a Lewis structure for the compound.
(c) Predict the shape of the molecules of the compound.
(d) What hybridization is consistent with the shape you predicted?
Consider nitrous acid, HNO 2 (HONO).
(a) Write a Lewis structure.
(b) What are the electron pair and molecular geometries of the internal oxygen and nitrogen atoms in the HNO 2 molecule?
(c) What is the hybridization on the internal oxygen and nitrogen atoms in HNO 2 ?
Strike-anywhere matches contain a layer of KClO 3 and a layer of P 4 S 3 . The heat produced by the friction of striking the match causes these two compounds to react vigorously, which sets fire to the wooden stem of the match. KClO 3 contains the ion. P 4 S 3 is an unusual molecule with the skeletal structure.
(a) Write Lewis structures for P 4 S 3 and the ion.
(b) Describe the geometry about the P atoms, the S atom, and the Cl atom in these species.
(c) Assign a hybridization to the P atoms, the S atom, and the Cl atom in these species.
(d) Determine the oxidation states and formal charge of the atoms in P 4 S 3 and the ion.
Identify the hybridization of each carbon atom in the following molecule. (The arrangement of atoms is given; you need to determine how many bonds connect each pair of atoms.)
Write Lewis structures for NF 3 and PF 5 . On the basis of hybrid orbitals, explain the fact that NF 3 , PF 3 , and PF 5 are stable molecules, but NF 5 does not exist.
In addition to NF 3 , two other fluoro derivatives of nitrogen are known: N 2 F 4 and N 2 F 2 . What shapes do you predict for these two molecules? What is the hybridization for the nitrogen in each molecule?
The bond energy of a C–C single bond averages 347 kJ mol −1 ; that of a triple bond averages 839 kJ mol −1 . Explain why the triple bond is not three times as strong as a single bond.
For the carbonate ion, draw all of the resonance structures. Identify which orbitals overlap to create each bond.
A useful solvent that will dissolve salts as well as organic compounds is the compound acetonitrile, H 3 CCN. It is present in paint strippers.
(a) Write the Lewis structure for acetonitrile, and indicate the direction of the dipole moment in the molecule.
(b) Identify the hybrid orbitals used by the carbon atoms in the molecule to form σ bonds.
(c) Describe the atomic orbitals that form the π bonds in the molecule. Note that it is not necessary to hybridize the nitrogen atom.
For the molecule allene, give the hybridization of each carbon atom. Will the hydrogen atoms be in the same plane or perpendicular planes?
Identify the hybridization of the central atom in each of the following molecules and ions that contain multiple bonds:
(a) ClNO (N is the central atom)
(b) CS 2
(c) Cl 2 CO (C is the central atom)
(d) Cl 2 SO (S is the central atom)
(e) SO 2 F 2 (S is the central atom)
(f) XeO 2 F 2 (Xe is the central atom)
(g) (Cl is the central atom)
Describe the molecular geometry and hybridization of the N, P, or S atoms in each of the following compounds.
(a) H 3 PO 4 , phosphoric acid, used in cola soft drinks
(b) NH 4 NO 3 , ammonium nitrate, a fertilizer and explosive
(c) S 2 Cl 2 , disulfur dichloride, used in vulcanizing rubber
(d) K 4 [O 3 POPO 3 ], potassium pyrophosphate, an ingredient in some toothpastes
For each of the following molecules, indicate the hybridization requested and whether or not the electrons will be delocalized:
(a) ozone (O 3 ) central O hybridization
(b) carbon dioxide (CO 2 ) central C hybridization
(c) nitrogen dioxide (NO 2 ) central N hybridization
(d) phosphate ion central P hybridization
For each of the following structures, determine the hybridization requested and whether the electrons will be delocalized:
(a) Hybridization of each carbon
(b) Hybridization of sulfur
(c) All atoms
Draw the orbital diagram for carbon in CO 2 showing how many carbon atom electrons are in each orbital.
Sketch the distribution of electron density in the bonding and antibonding molecular orbitals formed from two s orbitals and from two p orbitals.
How are the following similar, and how do they differ?
(a) σ molecular orbitals and π molecular orbitals
(b) ψ for an atomic orbital and ψ for a molecular orbital
(c) bonding orbitals and antibonding orbitals
If molecular orbitals are created by combining five atomic orbitals from atom A and five atomic orbitals from atom B combine, how many molecular orbitals will result?
Can a molecule with an odd number of electrons ever be diamagnetic? Explain why or why not.
Can a molecule with an even number of electrons ever be paramagnetic? Explain why or why not.
Why are bonding molecular orbitals lower in energy than the parent atomic orbitals?
Calculate the bond order for an ion with this configuration:
Explain why an electron in the bonding molecular orbital in the H 2 molecule has a lower energy than an electron in the 1 s atomic orbital of either of the separated hydrogen atoms.
Predict the valence electron molecular orbital configurations for the following, and state whether they will be stable or unstable ions.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
Determine the bond order of each member of the following groups, and determine which member of each group is predicted by the molecular orbital model to have the strongest bond.
(a) H 2 ,
(b) O 2 ,
(c) Li 2 , Be 2
(d) F 2 ,
(e) N 2 ,
For the first ionization energy for an N 2 molecule, what molecular orbital is the electron removed from?
Compare the atomic and molecular orbital diagrams to identify the member of each of the following pairs that has the highest first ionization energy (the most tightly bound electron) in the gas phase:
(a) H and H 2
(b) N and N 2
(c) O and O 2
(d) C and C 2
(e) B and B 2
Which of the period 2 homonuclear diatomic molecules are predicted to be paramagnetic?
A friend tells you that the 2 s orbital for fluorine starts off at a much lower energy than the 2 s orbital for lithium, so the resulting σ 2 s molecular orbital in F 2 is more stable than in Li 2 . Do you agree?
True or false: Boron contains 2 s 2 2 p 1 valence electrons, so only one p orbital is needed to form molecular orbitals.
What charge would be needed on F 2 to generate an ion with a bond order of 2?
Predict whether the MO diagram for S 2 would show s-p mixing or not.
Explain why is diamagnetic, while which has the same number of valence electrons, is paramagnetic.
Using the MO diagrams, predict the bond order for the stronger bond in each pair:
(a) B 2 or
(b) F 2 or
(c) O 2 or
(d) or


4.3: The Stages of Policy Development
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- Alison S. Burke, David Carter, Brian Fedorek, Tiffany Morey, Lore Rutz-Burri, & Shanell Sanchez
- Southern Oregon University via OpenOregon
The stages of policy development can generally be categorized into 5 general stages. U.S. policy development encompasses several stages. Most policy models generally include the following stages: (1) identifying the issue to be addressed by the proposed policy, (2) placement on the agenda, (3) formulation of the policy, (4) implementation of the policy, and (5) evaluation of the policy. This is similar to the community police response acronym SARA (scanning, analysis, response, and assessment) and uses some of the same techniques, but on a much bigger, national level.
Identifying the Problem and Agenda Setting
Identifying the problem involves addressing what is happening and why it is an issue. In criminal justice, this might look at the increase of opioid use and overdoses or acts of youth violence. Once the issue is identified, there can be a serious debate about the plans of the policy. Once it is decided what the policy will look like, it is placed on the agenda. This is perhaps the most politicized part of the process as it involves many different stakeholders. It involves identifying the legislative, regulatory, judicial, or other institutions responsible for policy adoption and formulation.
Formulation and Adoption
The next stage involved adopting the policy. Depending on the nature of the policy, this could involve a new law or an executive order.
Implementation of the Policy
Implementation is about moving forward, taking action, and spending money. It involves hiring new staff or additional police officers. This is where policies often stall because of the lack of funding. For example, a popular program in 1990, Weed and Seed, involved “weeding” out criminals (targeting arrest efforts) and “seeding” new programs (instituting after-school programs, drug treatment facilities, etc.). The weeding portion of the program was a great success, but the program ultimately failed because of a lack of funding to adequately seed new community programming. Funding is a major roadblock for proper implementation.
Evaluation
Finally, the evaluation examines the efficacy of the policy. There are three different types of evaluation: Impact, Process, and Cost-benefit analysis. Impact (outcome) evaluations focus on what changes after the introduction of the crime policy. [1] Changes in police patrol practices aimed at reducing the level of residential burglaries in an area are evaluated in terms of subsequent burglaries. The difficulty with impact evaluations is that changes in the crime rate are rarely, if ever, due to a single intervening variable. For example, after the implementation of curfew laws for juvenile offenders, juvenile crime decreased. Can we say that was because of curfew laws? The entire crime rate for America decreased at the same time. Attributing a single outcome based on a solitary intervention is problematic.
Process evaluations consider the implementation of a policy or program and involve determining the procedure used to implement the policy. These are detailed, descriptive accounts of the implementation of the policy including the goals of the program, who is involved, the level of training, the number of clients served, and changes to the program over time. [2] Unfortunately, process evaluations do not address the actual impact policy has on the crime problem, just what was done about a specific issue or who was involved. While this is indeed a limitation, it is essential to know the inner workings of a program or policy if it is to be replicated.
Cost-benefit evaluations , or analysis, seeks to determine if the costs of a policy are justified by the benefits accrued. A ubiquitous example of this would be an evaluation of the popular anti-drug D.A.R.E. program of the 1980s and 1990s. The D.A.R.E. program was a school-based prevention program aimed at preventing drug use among elementary school-aged children. Rigorous evaluations of the program show that it was ineffective and sometimes actually increased drug use in some youth. The cost of this program was roughly $1.3 billion dollars a year (about $173 to $268 per student per year) to implement nationwide (once all related expenses, such as police officer training and services, materials and supplies, school resources, etc., were factored in). [3] Using a cost-benefit analysis, is that a good use of money to support an ineffective program?
Policy formation is often a knee-jerk reaction to the current problem. Many policies are the result of grassroots efforts to change something in their communities. For example, let us pretend the issue is youth crime in our city. Kids are roaming the streets like packs of wild dogs, jeering at the elderly, and generally making us feel unsafe. A proposed policy might be to hold parents accountable for their child’s misbehavior. If parents are responsible, then they will take better care of their kids, right? Take, for example, Little Skippy.  He smokes, curses, and recently stole his neighbor’s car. Arrested after crashing into the drive-thru sign at the local Taco Bell, based on parental responsibility law, his mom and dad are to blame for his reckless driving fiasco. Let’s look at the policy process.
- How can this be instituted? Fine the parent? Sentence the parents to jail time? The policy needs to be a concrete solution to a problem. Many states use fines instead of jailing the parents. (Who’s to watch over the children if the parents are locked up?) Fines sound great. This will make sure parents take an active interest in their children because they do not want to have to pay money if their kid gets into trouble.
- Who needs to be involved in lobbying for this law? Legislators? Senators? Local police? Maybe even city officials, local school boards, and religious organizations. So it’s put on the agenda and gets moved onto a ballot for an official vote. The citizens who think their city needs to be tough on crime vote to approve this policy.
- Bam, it’s law. It is implemented and now parents of juveniles delinquents are charged fines. This actually is a law in nearly every state. In the 1990s, Silverton, Oregon, was a model for communities interested in imposing ordinances that hold parents accountable for their children’s behavior. In Silverton, parents can be fined up to $1,000 if their child is found carrying a gun, smoking cigarettes, or using illegal drugs. Parents who agree to attend parenting classes can avoid fines. Within the first two months after the law was passed in early 1995, seven parents were fined and many others registered for parenting classes.
Oregon has ORS 30.761 (2017), which states:
- In addition to any other remedy provided by law, the parent or parents of an unemancipated minor child shall be liable for actual damages to person or property caused by any tort intentionally or recklessly committed by such child. However, a parent who is not entitled to legal custody of the minor child at the time of the intentional or reckless tort shall not be liable for such damages.
- The legal obligation of the parent or parents of an unemancipated minor child to pay damages under this section shall be limited to not more than $7,500, payable to the same claimant, for one or more acts.
4. It is law, but is it effective? The evaluation stage of policy is critical. The goal is to curb youth crime and we might expect to see a decrease in the juvenile crime rate. However, charging parents fines for the misdeeds of their children actually increases recidivism! It’s true! A study of 1,167 youth in Pennsylvania found that the total amount of fines, fees and/or restitution significantly increased the likelihood of recidivism [4] . Justice system–imposed financial penalties increase the likelihood of recidivism in a sample of adolescent offenders [5] In particular, males, non-whites, and youth with prior dispositions and adjudicated with a drug or property offense were at an increased likelihood of recidivism associated with owing fines and fees (Piquero and Jennings, 2016). This is problematic as fees not only increase recidivism but also increase the likelihood of a “revolving door” juvenile justice system for minority youth.
In the end, what is law is not always effective and what is effective is not always law. This is where evidence-based practices come in.
- Lab, S. (2016). Crime Prevention: Approaches, Practices, and Evaluations (9th ed.). New York, NY: Routledge. ↵
- Lab, S. (2016). Crime Prevention: Approaches, Practices, and Evaluations (9th ed.). New York, NY: Routledge. ↵
- Shepard, E. (Winter 2001-2002) A new study finds. We wasted billions on D.A.R.E. Reconsider Quarterly, ↵
- Piquero and Jennings, 2016, Piquero, A. and Jennings, W.G. (2015) ↵
- Youth Violence and Juvenile Justice, 15 (3) p. 325-340). ↵


1.2: Native American
The selections of this section come from six tribes whose home‑ lands cover the majority of the United States’ eastern seaboard as well as regions in the midwest and southwest. The Micmac or Mi’kmaq tribe belonged to the Wabanaki Confederacy and occupied a region in southeastern Canada’s maritime provinces as well as parts of New York and New Jersey. One of the oldest political entities in the new world, the Haudenosaunee Confederacy were called the Iroquois by the French and the Five Nations by the English. The latter refers to the five tribes that made up the confederacy: the Cayuga, Mohawk, Oneida, Onondaga, and Seneca tribes. The name was changed to Six Nations when the Tuscarora tribe joined in the eighteenth century. Their territory covered the majority of New York with some inroads in southern Canada and northern Pennsylvania. Called the Delaware by Europeans, the Lenape tribe’s territory included what became New Jersey, eastern Pennsylvania, southeastern New York, northern Delaware, and a bit of southern Connecticut. The Cherokee tribe occupied the southeastern United States as far north as Kentucky and Virginia and as far south as Georgia and Alabama. The Winnebago, or the Ho‑chunks, lived in Wisconsin. Finally, the Zuni or the A:shimi were descendants of the ancient Anasazi and Mogollon cultures of the southwestern United States and occupied the area called New Mexico.
Image \(\PageIndex{1}\): Flag of the Wabanaki Confederacy
Missionaries and ethnologists were some of the first collectors of Native American tales. The missionaries often learned Native American languages and customs as a way to better proselytize the tribes, and some became at least as interested in these studies as in their religious missions. Moravian missionary John Heckewelder recorded the Lenape account of first contact before the American Revolution and published it early in the next century as part of the transactions of the American Philosophical Society, an outgrowth of the Federal era’s zeal for knowledge and scientific study. Baptist missionary Silas Rand ministered to the Micmac tribe and recorded the first contact story told to him by Micmac man Josiah Jeremy. A self‑taught linguist, Rand also published a Micmac dictionary. Toward the latter half of the nineteenth century, the developing field of ethnology—the analytic study of a culture’s customs, religious practices, and social structures—fueled the study of Native American culture. The Cherokee accounts recorded by James Mooney and the Zuni accounts recorded by Ruth Bunzel were first published as part of the annual reports produced by the Bureau of American Ethnology, a federal office in existence from 1879 to 1965 that authorized ethnological studies of tribes throughout America. Paul Radin—like Bunzel, a student of cultural anthropology pioneer Franz Boas—did his fieldwork for his doctorate among the Winnebago and there recorded the tribe’s trickster tales. While many of the accounts come from outsiders embedded for a time within tribes, some accounts were recorded by tribe members themselves. Though previously recounted by others, the Haudenosaunee creation story here is from Tuscarora tribal member David Cusick. A physician and artist, Cusick was one of the first Native American writers to preserve tribal history in his Sketches of the Ancient History of the Six Nations (1826). The Haudenosaunee Thanksgiving Address comes from University of Victoria professor Gerald Taiaiake Alfred, a member of the Kahnawake (Mohawk) tribe. Alfred has published several works about Native American culture in the early 21st century.
Image \(\PageIndex{2}\): Wampum Belt Commemorating the Iroquis Confederacy


5.5: Green's Theorem
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Learning Objectives
- Apply the circulation form of Green’s theorem.
- Apply the flux form of Green’s theorem.
- Calculate circulation and flux on more general regions.
In this section, we examine Green’s theorem, which is an extension of the Fundamental Theorem of Calculus to two dimensions. Green’s theorem has two forms: a circulation form and a flux form, both of which require region \(D\) in the double integral to be simply connected. However, we will extend Green’s theorem to regions that are not simply connected.
Put simply, Green’s theorem relates a line integral around a simply closed plane curve \(C\) and a double integral over the region enclosed by \(C\). The theorem is useful because it allows us to translate difficult line integrals into more simple double integrals, or difficult double integrals into more simple line integrals.
Extending the Fundamental Theorem of Calculus
Recall that the Fundamental Theorem of Calculus says that
\[\int_a^b F′(x)\,dx=F(b)−F(a). \nonumber \]
As a geometric statement, this equation says that the integral over the region below the graph of \(F′(x)\) and above the line segment \([a,b]\) depends only on the value of \(F\) at the endpoints \(a\) and \(b\) of that segment. Since the numbers \(a\) and \(b\) are the boundary of the line segment \([a,b]\), the theorem says we can calculate integral \(\int_a^b F′(x)\,dx\) based on information about the boundary of line segment \([a,b]\) (Figure \(\PageIndex{1}\)). The same idea is true of the Fundamental Theorem for Line Integrals:
\[\int_C \vecs \nabla f·d\vecs r=f(\vecs r(b))−f(\vecs r(a)). \nonumber \]
When we have a potential function (an “antiderivative”), we can calculate the line integral based solely on information about the boundary of curve \(C\).
Green’s theorem takes this idea and extends it to calculating double integrals. Green’s theorem says that we can calculate a double integral over region \(D\) based solely on information about the boundary of \(D\). Green’s theorem also says we can calculate a line integral over a simple closed curve \(C\) based solely on information about the region that \(C\) encloses. In particular, Green’s theorem connects a double integral over region \(D\) to a line integral around the boundary of \(D\).
Circulation Form of Green’s Theorem
The first form of Green’s theorem that we examine is the circulation form. This form of the theorem relates the vector line integral over a simple, closed plane curve \(C\) to a double integral over the region enclosed by \(C\). Therefore, the circulation of a vector field along a simple closed curve can be transformed into a double integral and vice versa.
GREEN’S THEOREM (CIRCULATION FORM)
Let \(D\) be an open, simply connected region with a boundary curve \(C\) that is a piecewise smooth, simple closed curve oriented counterclockwise (Figure \(\PageIndex{2}\)). Let \(\vecs F=⟨P,Q⟩\) be a vector field with component functions that have continuous partial derivatives on \(D\). Then,
\[ \begin{align} \oint_C \vecs F·d\vecs r =\oint_C P\,dx+Q\,dy \\[4pt] =\iint_D (Q_x−P_y)\,dA. \end{align} \nonumber \]
Notice that Green’s theorem can be used only for a two-dimensional vector field \(\vecs F\). If \(\vecs F\) is a three-dimensional field, then Green’s theorem does not apply. Since
\[\displaystyle \int_C P\,dx+Q\,dy=\int_C \vecs F·\vecs T\,ds \nonumber \]
this version of Green’s theorem is sometimes referred to as the tangential form of Green’s theorem.
The proof of Green’s theorem is rather technical, and beyond the scope of this text. Here we examine a proof of the theorem in the special case that \(D\) is a rectangle. For now, notice that we can quickly confirm that the theorem is true for the special case in which \(\vecs F=⟨P,Q⟩\) is conservative. In this case,
\[\oint_C P\,dx+Q\,dy=0 \nonumber \]
because the circulation is zero in conservative vector fields. \(\vecs F\) satisfies the cross-partial condition, so \(P_y=Q_x\). Therefore,
\[\iint_D (Q_x−P_y)\,dA=\int_D 0\,dA=0=\oint_C P\,dx+Q\,dy \nonumber \]
which confirms Green’s theorem in the case of conservative vector fields .
Proof
Let’s now prove that the circulation form of Green’s theorem is true when the region \(D\) is a rectangle. Let \(D\) be the rectangle \([a,b]×[c,d]\) oriented counterclockwise. Then, the boundary \(C\) of \(D\) consists of four piecewise smooth pieces \(C_1\), \(C_2\), \(C_3\), and \(C_4\) (Figure \(\PageIndex{3}\)). We parameterize each side of \(D\) as follows:
\(C_1: \vecs r_1(t)=⟨t,c⟩\), \(a≤t≤b\)
\(C_2: \vecs r_2(t)=⟨b,t⟩\), \(c≤t≤d\)
\(−C_3: \vecs r_3(t)=⟨t,d⟩\), \(a≤t≤b\)
\(−C_4: \vecs r_4(t)=⟨a,t⟩\), \(c≤t≤d\).
Then,
\[\begin{align*} \int_C \vecs F·d \vecs r &=\int_{C_1} \vecs F·d \vecs r+\int_{C_2} \vecs F·d \vecs r+\int_{C_3} \vecs F·d \vecs r+\int_{C_4} \vecs F·d \vecs r \\[4pt] &=\int_{C_1} \vecs F·d \vecs r+\int_{C_2} \vecs F·d \vecs r−\int_{−C_3} \vecs F·d \vecs r−\int_{−C_4} \vecs F·d \vecs r \\[4pt] &=\int_a^b \vecs F( \vecs r_1(t))· \vecs r_1'(t)\,dt+\int_c^d \vecs F( \vecs r_2(t))· \vecs r_2'(t)\,dt−\int_a^b \vecs F( \vecs r_3(t))· \vecs r_3'(t)\,dt−\int_c^d \vecs F( \vecs r_4(t))·\vecs r_4'(t)\,dt\\[4pt] &=\int_a^b P(t,c)\,dt+\int_c^dQ(b,t)\,dt−\int_a^bP(t,d)\,dt−\int_c^dQ(a,t)\,dt \\[4pt] &=\int_a^b(P(t,c)−P(t,d))\,dt+\int_c^d(Q(b,t)−Q(a,t))\,dt\\[4pt] &=−\int_a^b(P(t,d)−P(t,c))\,dt+\int_c^d(Q(b,t)−Q(a,t))\,dt. \end{align*}\]
By the Fundamental Theorem of Calculus,
\[P(t,d)−P(t,c)=\int_c^d \dfrac{\partial}{\partial y}P(t,y)dy \nonumber \]
and
\[Q(b,t)−Q(a,t)=\int_a^b \dfrac{\partial}{\partial x} Q(x,t)\,dx. \nonumber \]
Therefore,
\[−\int_a^b(P(t,d)−P(t,c))\,dt+\int_c^d(Q(b,t)−Q(a,t))\,dt=−\int_a^b\int_c^d \dfrac{\partial}{\partial y} P(t,y)\,dy\,dt+\int_c^d\int_a^b \dfrac{\partial}{\partial x}Q(x,t)\,dx\,dt. \nonumber \]
But,
\[\begin{align*} −\int_a^b\int_c^d \dfrac{\partial}{\partial y}P(t,y)\,dy\,dt+\int_c^d\int_a^b \dfrac{\partial}{\partial x}Q(x,t)\,dx\,dt &=−\int_a^b\int_c^d \dfrac{\partial}{\partial y}P(x,y)\,dy\,dx+\int_c^d\int_a^b \dfrac{\partial}{\partial x}Q(x,y)\,dx\,dy \\[4pt] &=\int_a^b\int_c^d(Q_x−P_y)\,dy\,dx\\[4pt] &=\iint_D(Q_x−P_y)\,dA. \end{align*}\]
Therefore, \(\displaystyle \int_C \vecs F\cdot d\vecs r=\iint_D(Q_x−P_y)\,dA\) and we have proved Green’s theorem in the case of a rectangle.
\(\square\)
To prove Green’s theorem over a general region \(D\), we can decompose \(D\) into many tiny rectangles and use the proof that the theorem works over rectangles. The details are technical, however, and beyond the scope of this text.
Example \(\PageIndex{1}\): Applying Green’s Theorem over a Rectangle
Calculate the line integral
\[\oint_C x^2y\,dx+(y−3)\,dy, \nonumber \]
where \(C\) is a rectangle with vertices \((1,1)\), \((4,1)\), \((4,5)\), and \((1,5)\) oriented counterclockwise.
Solution
Let \( \vecs F(x,y)=⟨P(x,y),Q(x,y)⟩=⟨x^2y,y−3⟩\). Then, \(Q_x(x,y)=0\) and \(P_y(x,y)=x^2\). Therefore, \(Q_x−P_y=−x^2\).
Let \(D\) be the rectangular region enclosed by \(C\) (Figure \(\PageIndex{4}\)). By Green’s theorem,
\[\begin{align*} \oint_C x^2y\,dx+(y−3)\,dy &=\iint_D (Q_x−P_y)\,dA \\[4pt] &=\iint_D −x^2 \,dA=\int_1^5\int_1^4−x^2\,dx\,dy \\[4pt] &= \int_1^5 -\frac{x^3}{3}\Bigg|_1^4\,dy\\[4pt] &= \int_1^5 \left( -\frac{64}{3} + \frac{1}{3} \right) \,dy \\[4pt] &=\int_1^5−21\,dy=−84.\end{align*}\]
Analysis
If we were to evaluate this line integral without using Green’s theorem, we would need to parameterize each side of the rectangle, break the line integral into four separate line integrals, and use the methods from the section titled Line Integrals to evaluate each integral. Furthermore, since the vector field here is not conservative, we cannot apply the Fundamental Theorem for Line Integrals. Green’s theorem makes the calculation much simpler.
Example \(\PageIndex{2}\): Applying Green’s Theorem to Calculate Work
Calculate the work done on a particle by force field
\[\vecs F(x,y)=⟨y+\sin x,e^y−x⟩ \nonumber \]
as the particle traverses circle \(x^2+y^2=4\) exactly once in the counterclockwise direction, starting and ending at point \((2,0)\).
Solution
Let \(C\) denote the circle and let \(D\) be the disk enclosed by \(C\). The work done on the particle is
\[W=\oint_C (y+\sin x)\,dx+(e^y−x)\,dy. \nonumber \]
As with Example \(\PageIndex{1}\), this integral can be calculated using tools we have learned, but it is easier to use the double integral given by Green’s theorem (Figure \(\PageIndex{5}\)).
Let \(\vecs F(x,y)=⟨P(x,y),Q(x,y)⟩=⟨y+\sin x,e^y−x⟩\). Then, \(Q_x=−1\) and \(P_y=1\). Therefore, \(Q_x−P_y=−2\).
By Green’s theorem,
\[\begin{align*} W &=\oint_C(y+\sin(x))dx+(e^y−x)\,dy \\[4pt] &=\iint_D (Q_x−P_y)\,dA \\[4pt] &=\iint_D−2\,dA \\[4pt] &=−2(area(D))=−2\pi (2^2)=−8\pi. \end{align*}\]
Exercise \(\PageIndex{2}\)
Use Green’s theorem to calculate line integral
\[\oint_C \sin(x^2)\,dx+(3x−y)\,dy. \nonumber \]
where \(C\) is a right triangle with vertices \((−1,2)\), \((4,2)\), and \((4,5)\) oriented counterclockwise.
- Hint
-
Transform the line integral into a double integral.
- Answer
-
\(\dfrac{45}{2}\)
In the preceding two examples, the double integral in Green’s theorem was easier to calculate than the line integral, so we used the theorem to calculate the line integral. In the next example, the double integral is more difficult to calculate than the line integral, so we use Green’s theorem to translate a double integral into a line integral.
Example \(\PageIndex{3}\): Applying Green’s Theorem over an Ellipse
Calculate the area enclosed by ellipse \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\) (Figure \(\PageIndex{6}\)).
Solution
Let \(C\) denote the ellipse and let \(D\) be the region enclosed by \(C\). Recall that ellipse \(C\) can be parameterized by
- \(x=a\cos t\),
- \(y=b \sin t\),
- \(0≤t≤2\pi\).
Calculating the area of \(D\) is equivalent to computing double integral \(\iint_D \,dA\). To calculate this integral without Green’s theorem, we would need to divide \(D\) into two regions: the region above the x -axis and the region below. The area of the ellipse is
\[\int_{−a}^a\int_0^{\sqrt{b^2−{(bx/a)}^2}} \,dy\,dx+\int_{−a}^{a} \int_{−\sqrt{b^2−{(bx/a)}^2}}^{0} \,dy\,dx. \nonumber \]
These two integrals are not straightforward to calculate (although when we know the value of the first integral, we know the value of the second by symmetry). Instead of trying to calculate them, we use Green’s theorem to transform \(\iint_D \,dA\) into a line integral around the boundary \(C\).
Consider vector field
\[\vecs F(x,y)=⟨P,Q⟩=⟨−\dfrac{y}{2},\dfrac{x}{2}⟩. \nonumber \]
Then, \(Q_x=\dfrac{1}{2}\) and \(P_y=−\dfrac{1}{2}\), and therefore \(Q_x−P_y=1\). Notice that \(\vecs F\) was chosen to have the property that \(Q_x−P_y=1\). Since this is the case, Green’s theorem transforms the line integral of \(\vecs F\) over \(C\) into the double integral of 1 over \(D\).
By Green’s theorem,
\[\begin{align*} \iint_D \,dA &=\iint_D (Q_x−P_y)\,dA \\[4pt] &=\int_C \vecs F\cdot d\vecs r=\dfrac{1}{2}\int_C −y\,dx+x\,dy \\[4pt] &=\dfrac{1}{2}\int_0^{2\pi}−b \sin t(−a\sin t)+a(\cos t)b\cos t\,dt \\[4pt] &=\dfrac{1}{2}\int_0^{2\pi} ab \cos^2 t+ab \sin^2 t\,dt \\[4pt] &=\dfrac{1}{2}\int_0^{2\pi} ab\,dt =\pi ab. \end{align*}\]
Therefore, the area of the ellipse is \(\pi ab\;\text{units}^2\).
In Example \(\PageIndex{3}\), we used vector field \(\vecs F(x,y)=⟨P,Q⟩=⟨−\dfrac{y}{2},\dfrac{x}{2}⟩\) to find the area of any ellipse. The logic of the previous example can be extended to derive a formula for the area of any region \(D\). Let \(D\) be any region with a boundary that is a simple closed curve \(C\) oriented counterclockwise. If \(\vecs F(x,y)=⟨P,Q⟩=⟨−\dfrac{y}{2},\dfrac{x}{2}⟩\), then \(Q_x−P_y=1\). Therefore, by the same logic as in Example \(\PageIndex{3}\),
\[ \text{area of} \; D=\iint_D \,dA=\dfrac{1}{2}\oint_C−y\,dx+x\,dy. \label{greenarea} \]
It’s worth noting that if \(\vecs F=⟨P,Q⟩\) is any vector field with \(Q_x−P_y=1\), then the logic of the previous paragraph works. So. Equation \ref{greenarea} is not the only equation that uses a vector field’s mixed partials to get the area of a region.
Exercise \(\PageIndex{3}\)
Find the area of the region enclosed by the curve with parameterization \(\vecs r(t)=⟨\sin t\cos t,\sin t⟩\), \(0≤t≤\pi\).
- Hint
-
Use Equation \ref{greenarea}.
- Answer
-
\(\dfrac{2}{3}\;\text{units}^2\).
- Solution
-
For convenience, let's use the same vector field we used in Example \(\PageIndex{3}\), where \(Q_x−P_y=1\).
\[\vecs F(x,y)=⟨P,Q⟩=⟨−\dfrac{y}{2},\dfrac{x}{2}⟩. \nonumber \]
For the curve \(r(t)=⟨\sin t\cos t,\sin t⟩\) with \(0≤t≤\pi\), we have
\[\begin{align*} \vecs r'(t) &= \langle \cos^2t - \sin^2t, \cos t \rangle \text{ and} \\[5pt]
\vecs F(t) &= \left\langle -\frac{\sin t}{2}, \frac{\sin t \cos t}{2} \right\rangle \end{align*}\]
By Green’s theorem,\[\begin{align*} \text{Area of the region }D =\iint_D \,dA &=\iint_D (Q_x−P_y)\,dA \\[4pt] &=\int_C \vecs F\cdot d\vecs r=\dfrac{1}{2}\int_C −y\,dx+x\,dy \\[4pt] &=\dfrac{1}{2}\int_0^{\pi}\big(−\sin t(\cos^2t - \sin^2t)+\sin t \cos^2 t \big)\,dt \\[4pt]
&=\dfrac{1}{2}\int_0^{\pi} \big(−\sin t\cos^2t + \sin t\sin^2 t + \sin t\cos^2 t \big) \,dt \\[4pt]
&=\dfrac{1}{2}\int_0^{\pi} \big(\cancel{−\sin t\cos^2t} + \sin t(1 - \cos^2 t) + \cancel{\sin t\cos^2 t} \big) \,dt \\[4pt]
&=\dfrac{1}{2}\int_0^{\pi} \big( \sin t - \cos^2 t\sin t \big) \,dt \\[4pt]
&= \frac{1}{2} \left[ -\cos t + \frac{\cos^3 t}{3} \right]\Bigg |_0^{\pi}\\[4pt]
&= \frac{1}{2} \left[ -\cos \pi + \frac{\cos^3 \pi}{3} + \cos 0 - \frac{\cos^3 0}{3} \right]\\[4pt]
&= \frac{1}{2} \left[ 1 - \frac{1}{3} + 1 - \frac{1}{3} \right]\\[4pt]
&= \frac{1}{2} \left[ 2 - \frac{2}{3} \right]\\[4pt]
&= \frac{1}{2} \left[ \frac{6}{3} - \frac{2}{3} \right]\\[4pt]
&= \frac{1}{2} \left[ \frac{4}{3} \right]\\[4pt]
&= \frac{2}{3} \;\text{units}^2 \end{align*}\]Therefore the area of this region is \(\dfrac{2}{3}\;\text{units}^2\).
Flux Form of Green’s Theorem
The circulation form of Green’s theorem relates a double integral over region \(D\) to line integral \(\displaystyle \oint_C \vecs F·\vecs Tds\), where \(C\) is the boundary of \(D\). The flux form of Green’s theorem relates a double integral over region \(D\) to the flux across boundary \(C\). The flux of a fluid across a curve can be difficult to calculate using the flux line integral. This form of Green’s theorem allows us to translate a difficult flux integral into a double integral that is often easier to calculate.
GREEN’S THEOREM (FLUX FORM)
Let \(D\) be an open, simply connected region with a boundary curve \(C\) that is a piecewise smooth, simple closed curve that is oriented counterclockwise (Figure \(\PageIndex{7}\)). Let \(\vecs F=⟨P,Q⟩\) be a vector field with component functions that have continuous partial derivatives on an open region containing \(D\). Then,
\[\oint_C \vecs F·\vecs N\,ds=\iint_D P_x+Q_y\,dA. \label{GreenN} \]
Because this form of Green’s theorem contains unit normal vector \(\vecs N\), it is sometimes referred to as the normal form of Green’s theorem.
Proof
Recall that \(\displaystyle \oint_C \vecs F·\vecs N\,ds=\oint_C −Q\,dx+P\,dy\). Let \(M=−Q\) and \(N=P\). By the circulation form of Green’s theorem,
\[\begin{align*} \oint_C−Q\,dx+P\,dy &=\oint_C M\,dx+N\,dy\\[4pt] &=\iint_D N_x−M_y \,dA\\[4pt] &=\iint_D P_x−{(−Q)}_y \,dA\\[4pt] &=\iint_D P_x+Q_y \,dA. \end{align*}\]
\(\square\)
Example \(\PageIndex{4A}\): Applying Green’s Theorem for Flux across a Circle
Let \(C\) be a circle of radius \(r\) centered at the origin (Figure \(\PageIndex{8}\)) and let \(\vecs F(x,y)=⟨x,y⟩\). Calculate the flux across \(C\).
Solution
Let \(D\) be the disk enclosed by \(C\). The flux across \(C\) is \(\displaystyle \oint_C \vecs F·\vecs N\,ds\). We could evaluate this integral using tools we have learned, but Green’s theorem makes the calculation much more simple. Let \(P(x,y)=x\) and \(Q(x,y)=y\) so that \(\vecs F=⟨P,Q⟩\). Note that \(P_x=1=Q_y\), and therefore \(P_x+Q_y=2\). By Green’s theorem,
\[\int_C \vecs F\cdot\vecs N\,ds=\iint_D 2\,dA=2\iint_D \,dA. \nonumber \]
Since \(\displaystyle \iint_D \,dA\) is the area of the circle, \(\displaystyle \iint_D \,dA=\pi r^2\). Therefore, the flux across \(C\) is \(2\pi r^2\).
Example \(\PageIndex{4B}\): Applying Green’s Theorem for Flux across a Triangle
Let \(S\) be the triangle with vertices \((0,0)\), \((1,0)\), and \((0,3)\) oriented clockwise (Figure \(\PageIndex{9}\)). Calculate the flux of \(\vecs F(x,y)=⟨P(x,y),Q(x,y)⟩=⟨x^2+e^y,x+y⟩\) across \(S\).
Solution
To calculate the flux without Green’s theorem, we would need to break the flux integral into three line integrals, one integral for each side of the triangle. Using Green’s theorem to translate the flux line integral into a single double integral is much more simple.
Let \(D\) be the region enclosed by \(S\). Note that \(P_x=2x\) and \(Q_y=1\); therefore, \(P_x+Q_y=2x+1\). Green’s theorem applies only to simple closed curves oriented counterclockwise, but we can still apply the theorem because \(\displaystyle \oint_C \vecs F·\vecs N\,ds=−\oint_{−S} \vecs F·\vecs N\,ds\) and \(−S\) is oriented counterclockwise. By Green’s theorem, the flux is
\[\begin{align*} \oint_C \vecs F·\vecs N\,ds &= \oint_{−S} \vecs F·\vecs N\,ds\\[4pt] &=−\iint_D (P_x+Q_y)\,dA \\[4pt] &=−\iint_D (2x+1)\,dA.\end{align*}\]
Notice that the top edge of the triangle is the line \(y=−3x+3\). Therefore, in the iterated double integral, the \(y\)-values run from \(y=0\) to \(y=−3x+3\), and we have
\[\begin{align*} −\iint_D (2x+1)\,dA &= −\int_0^1\int_0^{−3x+3}(2x+1)\,dy\,dx \\[4pt]
&=−\int_0^1(2x+1)y\Big|_0^{-3x+3}\,dx \\[4pt]
&=−\int_0^1(2x+1)(−3x+3)\,dx \\[4pt]
&=−\int_0^1(−6x^2+3x+3)\,dx\\[4pt]
&=−{\left[−2x^3+\dfrac{3x^2}{2}+3x\right]}\Bigg|_0^1 \\[4pt]
&=−\dfrac{5}{2}. \end{align*}\]
Exercise \(\PageIndex{4}\)
Calculate the flux of \(\vecs F(x,y)=⟨x^3,y^3⟩\) across a unit circle oriented counterclockwise.
- Hint
-
Apply Green’s theorem and use polar coordinates.
- Answer
-
\(\dfrac{3\pi}{2}\)
Example \(\PageIndex{5}\): Applying Green’s Theorem for Water Flow across a Rectangle
Water flows from a spring located at the origin. The velocity of the water is modeled by vector field \(\vecs v(x,y)=⟨5x+y,x+3y⟩\) m/sec. Find the amount of water per second that flows across the rectangle with vertices \((−1,−2)\), \((1,−2)\), \((1,3)\),and \((−1,3)\), oriented counterclockwise (Figure \(\PageIndex{10}\)).
Solution
Let \(C\) represent the given rectangle and let \(D\) be the rectangular region enclosed by \(C\). To find the amount of water flowing across \(C\), we calculate flux \(\displaystyle \int_C \vecs v\cdot d\vecs r\). Let \(P(x,y)=5x+y\) and \(Q(x,y)=x+3y\) so that \(\vecs v=⟨P,Q⟩\). Then, \(P_x=5\) and \(Q_y=3\). By Green’s theorem,
\[\begin{align*} \int_C \vecs v\cdot d\vecs r &=\iint_D (P_x+Q_y)\,dA \\ &=\iint_D 8\,dA \\ &=8(area\space of\space D)=80. \end{align*}\]
Therefore, the water flux is 80 m 2 /sec.
Recall that if vector field \(\vecs F\) is conservative, then \(\vecs F\) does no work around closed curves—that is, the circulation of \(\vecs F\) around a closed curve is zero. In fact, if the domain of \(\vecs F\) is simply connected, then \(\vecs F\) is conservative if and only if the circulation of \(\vecs F\) around any closed curve is zero. If we replace “circulation of \(\vecs F\)” with “flux of \(\vecs F\),” then we get a definition of a source-free vector field. The following statements are all equivalent ways of defining a source-free field \(\vecs F=⟨P,Q⟩\) on a simply connected domain (note the similarities with properties of conservative vector fields):
- The flux \( \displaystyle \oint_C \vecs F·\vecs N\,ds\) across any closed curve \(C\) is zero.
- If \(C_1\) and \(C_2\) are curves in the domain of \(\vecs F\) with the same starting points and endpoints, then \(\displaystyle \int_{C_1} \vecs F·\vecs N\,ds=\int_{C_2} \vecs F·\vecs N\,ds\). In other words, flux is independent of path.
- There is a stream function \(g(x,y)\) for \(\vecs F\). A stream function for \(\vecs F=⟨P,Q⟩\) is a function g such that \(P=g_y\) and \(Q=−g_x\).Geometrically, \(\vecs F=\langle a,b\rangle\) is tangential to the level curve of \(g\) at \((a,b)\). Since the gradient of \(g\) is perpendicular to the level curve of \(g\) at \((a,b)\), stream function \(g\) has the property \(\vecs F(a,b)\cdot\vecs\nabla g(a,b)=0\) for any point \((a,b)\) in the domain of \(g\). (Stream functions play the same role for source-free fields that potential functions play for conservative fields.)
- \(P_x+Q_y=0\)
Example \(\PageIndex{6}\): Finding a Stream Function
Verify that rotation vector field \(\vecs F(x,y)=⟨y,−x⟩\) is source free, and find a stream function for \(\vecs F\).
Solution
Note that the domain of \(\vecs F\) is all of \(ℝ^2\), which is simply connected. Therefore, to show that \(\vecs F\) is source free, we can show any of items 1 through 4 from the previous list to be true. In this example, we show that item 4 is true. Let \(P(x,y)=y\) and \(Q(x,y)=−x\). Then \(P_x=0=Q_y\), and therefore \(P_x+Q_y=0\). Thus, \(\vecs F\) is source free.
To find a stream function for \(\vecs F\), proceed in the same manner as finding a potential function for a conservative field. Let \(g\) be a stream function for \(\vecs F\). Then \(g_y=y\), which implies that
\(g(x,y)=\dfrac{y^2}{2}+h(x)\).
Since \(−g_x=Q=−x\), we have \(h′(x)=x\). Therefore,
\(h(x)=\dfrac{x^2}{2}+C\).
Letting \(C=0\) gives stream function
\(g(x,y)=\dfrac{x^2}{2}+\dfrac{y^2}{2}\).
To confirm that \(g\) is a stream function for \(\vecs F\), note that \(g_y=y=P\) and \(−g_x=−x=Q\).
Notice that source-free rotation vector field \(\vecs F(x,y)=⟨y,−x⟩\) is perpendicular to conservative radial vector field \(\vecs \nabla g=⟨x,y⟩\) (Figure \(\PageIndex{11}\)).
Exercise \(\PageIndex{6}\)
Find a stream function for vector field \(\vecs F(x,y)=⟨x \sin y,\cos y⟩\).
- Hint
-
Follow the outline provided in the previous example.
- Answer
-
\(g(x,y)=−x\cos y\)
Vector fields that are both conservative and source free are important vector fields. One important feature of conservative and source-free vector fields on a simply connected domain is that any potential function \(f\) of such a field satisfies Laplace’s equation \(f_{xx}+f_{yy}=0\). Laplace’s equation is foundational in the field of partial differential equations because it models such phenomena as gravitational and magnetic potentials in space, and the velocity potential of an ideal fluid. A function that satisfies Laplace’s equation is called a harmonic function. Therefore any potential function of a conservative and source-free vector field is harmonic.
To see that any potential function of a conservative and source-free vector field on a simply connected domain is harmonic, let \(f\) be such a potential function of vector field \(\vecs F=⟨P,Q⟩\). Then, \(f_x=P\) and \(f_x=Q\) because \(\vecs \nabla f=\vecs F\). Therefore, \(f_{xx}=P_x\) and \(f_{yy}=Q_y\). Since \(\vecs F\) is source free, \(f_{xx}+f_{yy}=P_x+Q_y=0\), and we have that \(f\) is harmonic.
Example \(\PageIndex{7}\): Satisfying Laplace’s Equation
For vector field \(\vecs F(x,y)=⟨e^x\sin y,e^x\cos y⟩\), verify that the field is both conservative and source free, find a potential function for \(\vecs F\), and verify that the potential function is harmonic.
Solution
Let \(P(x,y)=e^x\sin y\) and \(Q(x,y)=e^x \cos y\). Notice that the domain of \(\vecs F\) is all of two-space, which is simply connected. Therefore, we can check the cross-partials of \(\vecs F\) to determine whether \(\vecs F\) is conservative. Note that \(P_y=e^x \cos y=Q_x\), so \(\vecs F\) is conservative. Since \(P_x=e^x \sin y\) and \(Q_y=e^x \sin y\), \(P_x+Q_y=0\) and the field is source free.
To find a potential function for \(\vecs F\), let \(f\) be a potential function. Then, \(\vecs \nabla f=\vecs F\), so \(f_x(x,y)=e^x \sin y\). Integrating this equation with respect to x gives \(f(x,y)=e^x \sin y+h(y)\). Since \(f_y(x,y)=e^x \cos y\), differentiating \(f\) with respect to y gives \(e^x\cos y=e^x\cos y+h′(y)\). Therefore, we can take \(h(y)=0\), and \(f(x,y)=e^x\sin y\) is a potential function for \(f\).
To verify that \(f\) is a harmonic function, note that \(f_{xx}(x,y)=\dfrac{\partial}{\partial x}(e^x\sin y)=e^x \sin y\) and
\(f_{yy}(x,y)=\dfrac{\partial}{\partial x}(e^x\cos y)=−e^x\sin y\). Therefore, \(f_{xx}+f_{yy}=0\), and \(f\) satisfies Laplace’s equation.
Exercise \(\PageIndex{7}\)
Is the function \(f(x,y)=e^{x+5y}\) harmonic?
- Hint
-
Determine whether the function satisfies Laplace’s equation.
- Answer
-
No
Green’s Theorem on General Regions
Green’s theorem, as stated, applies only to regions that are simply connected—that is, Green’s theorem as stated so far cannot handle regions with holes. Here, we extend Green’s theorem so that it does work on regions with finitely many holes (Figure \(\PageIndex{12}\)).
Before discussing extensions of Green’s theorem, we need to go over some terminology regarding the boundary of a region. Let \(D\) be a region and let \(C\) be a component of the boundary of \(D\). We say that \(C\) is positively oriented if, as we walk along \(C\) in the direction of orientation, region \(D\) is always on our left. Therefore, the counterclockwise orientation of the boundary of a disk is a positive orientation, for example. Curve \(C\) is negatively oriented if, as we walk along \(C\) in the direction of orientation, region \(D\) is always on our right. The clockwise orientation of the boundary of a disk is a negative orientation, for example.
Let \(D\) be a region with finitely many holes (so that \(D\) has finitely many boundary curves), and denote the boundary of \(D\) by \(\partial D\) (Figure \(\PageIndex{13}\)). To extend Green’s theorem so it can handle \(D\), we divide region \(D\) into two regions, \(D_1\) and \(D_2\) (with respective boundaries \(\partial D_1\) and \(\partial D_2\)), in such a way that \(D=D_1\cup D_2\) and neither \(D_1\) nor \(D_2\) has any holes (Figure \(\PageIndex{13}\)).
Assume the boundary of \(D\) is oriented as in the figure, with the inner holes given a negative orientation and the outer boundary given a positive orientation. The boundary of each simply connected region \(D_1\) and \(D_2\) is positively oriented. If \(\vecs F\) is a vector field defined on \(D\), then Green’s theorem says that
\[\begin{align} \oint_{\partial D} \vecs F·d\vecs{r} &=\oint_{\partial D_1}\vecs F·d\vecs{r}+\oint_{\partial D_2}\vecs F·d\vecs{r} \\ &=\iint_{D_1}Q_x−P_y\,dA+\iint_{D_2}Q_x−P_y\,dA \\ &=\iint_D (Q_x−P_y)\,dA.\end{align} \nonumber \]
Therefore, Green’s theorem still works on a region with holes.
To see how this works in practice, consider annulus \(D\) in Figure \(\PageIndex{14}\) and suppose that \(F=⟨P,Q⟩\) is a vector field defined on this annulus. Region \(D\) has a hole, so it is not simply connected. Orient the outer circle of the annulus counterclockwise and the inner circle clockwise (Figure \(\PageIndex{14}\)) so that, when we divide the region into \(D_1\) and \(D_2\), we are able to keep the region on our left as we walk along a path that traverses the boundary. Let \(D_1\) be the upper half of the annulus and \(D_2\) be the lower half. Neither of these regions has holes, so we have divided \(D\) into two simply connected regions.
We label each piece of these new boundaries as \(P_i\) for some \(i\), as in Figure \(\PageIndex{14}\). If we begin at \(P\) and travel along the oriented boundary, the first segment is \(P_1\), then \(P_2\), \(P_3\), and \(P_4\). Now we have traversed \(D_1\) and returned to \(P\). Next, we start at \(P\) again and traverse \(D_2\). Since the first piece of the boundary is the same as \(P_4\) in \(D_1\), but oriented in the opposite direction, the first piece of \(D_2\) is \(−P_4\). Next, we have \(P_5\), then \(−P_2\), and finally \(P_6\).
Figure \(\PageIndex{14}\) shows a path that traverses the boundary of \(D\). Notice that this path traverses the boundary of region \(D_1\), returns to the starting point, and then traverses the boundary of region \(D_2\). Furthermore, as we walk along the path, the region is always on our left. Notice that this traversal of the \(P_i\) paths covers the entire boundary of region \(D\). If we had only traversed one portion of the boundary of \(D\), then we cannot apply Green’s theorem to \(D\).
The boundary of the upper half of the annulus, therefore, is \(P_1\cup P_2\cup P_3\cup P_4\) and the boundary of the lower half of the annulus is \(−P_4\cup P_5\cup −P_2\cup P_6\). Then, Green’s theorem implies
\[\begin{align} \oint_{\partial D}\vecs F·d\vecs{r} &=\int_{P_1}\vecs F·d\vecs{r}+\int_{P_2}\vecs F·d\vecs{r}+\int_{P_3}\vecs F·d\vecs{r}+\int_{P_4}\vecs F·d\vecs{r}+\int_{−P_4}\vecs F·d\vecs{r}+\int_{P_5}\vecs F·d\vecs{r}+\int_{−P_2}\vecs F·d\vecs{r}+\int_{P_6}\vecs F·d\vecs{r} \\ &=\int_{P_1}\vecs F·d\vecs{r}+\int_{P_2}\vecs F·d\vecs{r}+\int_{P_3}\vecs F·d\vecs{r}+\int_{P_4}\vecs F·d\vecs{r}+\int_{P_4}\vecs F·d\vecs{r}+\int_{P_5}\vecs F·d\vecs{r}+\int_{−P_2}\vecs F·d\vecs{r}+\int_{P_6}\vecs F·d\vecs{r} \\ &=\int_{P_1}\vecs F·d\vecs{r}+\int_{P_3}\vecs F·d\vecs{r}+\int_{P_5}\vecs F·d\vecs{r}+\int_{P_6}\vecs F·d\vecs{r} \\ &=\oint_{\partial D_1}\vecs F·d\vecs{r}+\oint_{\partial D_2}\vecs F·d\vecs{r}\\ &=\iint_{D_1}(Q_x−P_y)\,dA+\iint_{D_2}(Q_x−P_y)\,dA \\ &=\iint_D(Q_x−P_y)\,dA. \end{align} \nonumber \]
Therefore, we arrive at the equation found in Green’s theorem—namely,
\[\oint_{\partial D}\vecs F·d\vecs{r}=\iint_D (Q_x−P_y)\,dA. \nonumber \]
The same logic implies that the flux form of Green’s theorem can also be extended to a region with finitely many holes:
\[\oint_C \vecs F·\vecs N\,ds=\iint_D (P_x+Q_y)\,dA. \nonumber \]
Example \(\PageIndex{8A}\): Using Green’s Theorem on a Region with Holes
Calculate the integral
\[\oint_{\partial D}(\sin x−\dfrac{y^3}{3})dx+(\dfrac{y^3}{3}+\sin y)dy, \nonumber \]
where \(D\) is the annulus given by the polar inequalities \(1≤r≤2\), \(0≤\theta≤2\pi\).
Solution
Although \(D\) is not simply connected, we can use the extended form of Green’s theorem to calculate the integral. Since the integration occurs over an annulus, we convert to polar coordinates:
\[\begin{align*} \oint_{\partial D}(\sin x−\dfrac{y^3}{3})\,dx+(\dfrac{x^3}{3}+\sin y)\,dy &=\iint_D (Q_x−P_y)\,dA \\ &=\iint_D (x^2+y^2)\,dA\\ &=\int_0^{2\pi}\int_1^2 r^3\,drd\theta=\int_0^{2\pi} \dfrac{15}{4}\,d\theta \\ &=\dfrac{15\pi}{2}. \end{align*}\]
Example \(\PageIndex{8B}\): Using the Extended Form of Green’s Theorem
Let \(\vecs F=⟨P,Q⟩=\left\langle\dfrac{y}{x^2+y^2},-\dfrac{x}{x^2+y^2}\right\rangle\) and let \(C\) be any simple closed curve in a plane oriented counterclockwise. What are the possible values of \(\displaystyle \oint_C \vecs F·d\vecs{r}\)?
Solution
We use the extended form of Green’s theorem to show that \(\displaystyle \oint_C \vecs F·d\vecs{r}\) is either \(0\) or \(−2\pi\)—that is, no matter how crazy curve \(C\) is, the line integral of \(\vecs F\) along \(C\) can have only one of two possible values. We consider two cases: the case when \(C\) encompasses the origin and the case when \(C\) does not encompass the origin.
Case 1: C Does Not Encompass the Origin
In this case, the region enclosed by \(C\) is simply connected because the only hole in the domain of \(\vecs F\) is at the origin. We showed in our discussion of cross-partials that \(\vecs F\) satisfies the cross-partial condition. If we restrict the domain of \(\vecs F\) just to \(C\) and the region it encloses, then \(\vecs F\) with this restricted domain is now defined on a simply connected domain. Since \(\vecs F\) satisfies the cross-partial property on its restricted domain, the field \(\vecs F\) is conservative on this simply connected region and hence the circulation \(\displaystyle \oint_C \vecs F·d\vecs{r}\) is zero.
Case 2: C Does Encompass the Origin
In this case, the region enclosed by \(C\) is not simply connected because this region contains a hole at the origin. Let \(C_1\) be a circle of radius a centered at the origin so that \(C_1\) is entirely inside the region enclosed by \(C\) (Figure \(\PageIndex{15}\)). Give \(C_1\) a clockwise orientation.
Let \(D\) be the region between \(C_1\) and \(C\), and \(C\) is orientated counterclockwise. By the extended version of Green’s theorem,
\[\begin{align*} \int_C \vecs F·d\vecs{r}+\int_{C_1}\vecs F·d\vecs{r} &=\iint_D Qx_−P_y \,dA \\[4pt] &=\iint_D−\dfrac{y^2−x^2}{{(x^2+y^2)}^2}+\dfrac{y^2−x^2}{{(x^2+y^2)}^2}dA \\[4pt] &=0, \end{align*}\]
and therefore
\[\int_C \vecs F·d\vecs{r}=−\int_{C_1} \vecs F·d\vecs{r}. \nonumber \]
Since \(C_1\) is a specific curve, we can evaluate \(\displaystyle \int_{C_1}\vecs F·d\vecs{r}\). Let
\[ x=a\cos t, \;\; y=a\sin t, \;\; 0≤t≤2\pi \nonumber \]
be a parameterization of \(C_1\). Then,
\[\begin{align*} \int_{C_1}\vecs F·d\vecs{r} &=\int_0^{2\pi} \vecs F(\vecs r(t))·\vecs r′(t)\,dt \\[4pt] &=\int_0^{2\pi} ⟨−\dfrac{\sin(t)}{a},−\dfrac{\cos(t)}{a}⟩·⟨−a\sin(t),−a\cos(t)⟩\,dt \\[4pt] &=\int_0^{2\pi}{\sin}^2(t)+{\cos}^2(t)\,dt \\[4pt] &=\int_0^{2\pi}\,dt=2\pi. \end{align*}\]
Therefore, \(\displaystyle \int_C \vecs F·d\vecs r=−2\pi\).
Exercise \(\PageIndex{8}\)
Calculate integral \(\displaystyle \oint_{\partial D}\vecs F·d\vecs{r}\), where \(D\) is the annulus given by the polar inequalities \(2≤r≤5\), \(0≤\theta≤2\pi\), and \(\vecs F(x,y)=⟨x^3,5x+e^y\sin y⟩\).
- Hint
-
Use the extended version of Green’s theorem.
 The brain has a tumor (Figure \(\PageIndex{16}\)). How large is the tumor? To be precise, what is the area of the red region? The red cross-section of the tumor has an irregular shape, and therefore it is unlikely that you would be able to find a set of equations or inequalities for the region and then be able to calculate its area by conventional means. You could approximate the area by chopping the region into tiny squares (a Riemann sum approach), but this method always gives an answer with some error.
Instead of trying to measure the area of the region directly, we can use a device called a rolling planimeter to calculate the area of the region exactly, simply by measuring its boundary. In this project you investigate how a planimeter works, and you use Green’s theorem to show the device calculates area correctly.
A rolling planimeter is a device that measures the area of a planar region by tracing out the boundary of that region (Figure \(\PageIndex{17}\)). To measure the area of a region, we simply run the tracer of the planimeter around the boundary of the region. The planimeter measures the number of turns through which the wheel rotates as we trace the boundary; the area of the shape is proportional to this number of wheel turns. We can derive the precise proportionality equation using Green’s theorem. As the tracer moves around the boundary of the region, the tracer arm rotates and the roller moves back and forth (but does not rotate).
Let \(C\) denote the boundary of region \(D\), the area to be calculated. As the tracer traverses curve \(C\), assume the roller moves along the y -axis (since the roller does not rotate, one can assume it moves along a straight line). Use the coordinates \((x,y)\) to represent points on boundary \(C\), and coordinates \((0,Y)\) to represent the position of the pivot. As the planimeter traces \(C\), the pivot moves along the y -axis while the tracer arm rotates on the pivot.
Watch a short animation of a planimeter in action.
Begin the analysis by considering the motion of the tracer as it moves from point \((x,y)\) counterclockwise to point \((x+dx,y+dy)\) that is close to \((x,y)\) (Figure \(\PageIndex{18}\)). The pivot also moves, from point \((0,Y)\) to nearby point \((0,Y+dY)\). How much does the wheel turn as a result of this motion? To answer this question, break the motion into two parts. First, roll the pivot along the y -axis from \((0,Y)\) to \((0,Y+dY)\) without rotating the tracer arm. The tracer arm then ends up at point \((x,y+dY)\) while maintaining a constant angle \(\phi\) with the x -axis. Second, rotate the tracer arm by an angle \(d\theta\) without moving the roller. Now the tracer is at point \((x+dx,y+dy)\). Let ll be the distance from the pivot to the wheel and let \(L\) be the distance from the pivot to the tracer (the length of the tracer arm).
- Explain why the total distance through which the wheel rolls the small motion just described is \(\sin \phi dY+ld\theta=\dfrac{x}{L}dY+ld\theta\).
- Show that \(\oint_C d\theta=0\).
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Use step 2 to show that the total rolling distance of the wheel as the tracer traverses curve \(C\) is
Total wheel roll \(=\dfrac{1}{L}\oint_C xdY\).
Now that you have an equation for the total rolling distance of the wheel, connect this equation to Green’s theorem to calculate area \(D\) enclosed by \(C\). - Show that \(x^2+(y−Y)^2=L^2\).
- Assume the orientation of the planimeter is as shown in Figure \(\PageIndex{18}\). Explain why \(Y≤y\), and use this inequality to show there is a unique value of \(Y\) for each point \((x,y)\): \(Y=y=\sqrt{L^2−x^2}\).
- Use step 5 to show that \(dY=dy+\dfrac{x}{L^2−x^2}dx.\)
- Use Green’s theorem to show that \(\displaystyle \oint_C \dfrac{x}{L^2−x^2}dx=0\).
-
Use step 7 to show that the total wheel roll is
\[\text{Total wheel roll}\quad =\quad 1L\oint_C x\,dy. \nonumber \]
It took a bit of work, but this equation says that the variable of integration Y in step 3 can be replaced with y .
- Use Green’s theorem to show that the area of \(D\) is \(\displaystyle \oint_C x\,dy\). The logic is similar to the logic used to show that the area of \(\displaystyle D=12\oint_C −y\,dx+x\,dy\).
- Conclude that the area of \(D\) equals the length of the tracer arm multiplied by the total rolling distance of the wheel.
You now know how a planimeter works and you have used Green’s theorem to justify that it works. To calculate the area of a planar region \(D\), use a planimeter to trace the boundary of the region. The area of the region is the length of the tracer arm multiplied by the distance the wheel rolled.
Key Concepts
- Green’s theorem relates the integral over a connected region to an integral over the boundary of the region. Green’s theorem is a version of the Fundamental Theorem of Calculus in one higher dimension.
- Green’s Theorem comes in two forms: a circulation form and a flux form. In the circulation form, the integrand is \(\vecs F·\vecs T\). In the flux form, the integrand is \(\vecs F·\vecs N\).
- Green’s theorem can be used to transform a difficult line integral into an easier double integral, or to transform a difficult double integral into an easier line integral.
- A vector field is source free if it has a stream function. The flux of a source-free vector field across a closed curve is zero, just as the circulation of a conservative vector field across a closed curve is zero.
Key Equations
-
Green’s theorem, circulation form
\(\displaystyle ∮_C P\,dx+Q\,dy=∬_D Q_x−P_y\,dA\), where \(C\) is the boundary of \(D\) -
Green’s theorem, flux form
\(\displaystyle ∮_C\vecs F·\vecs N\,ds=∬_D P_x+Q_y\,dA\), where \(C\) is the boundary of \(D\) -
Green’s theorem, extended version
\(\displaystyle ∮_{\partial D}\vecs F·d\vecs{r}=∬_D Q_x−P_y\,dA\)
Glossary
- Green’s theorem
- relates the integral over a connected region to an integral over the boundary of the region
- stream function
- if \(\vecs F=⟨P,Q⟩\) is a source-free vector field, then stream function \(g\) is a function such that \(P=g_y\) and \(Q=−g_x\)
Contributions
Paul Seeburger created a corrected image and the worked out solution for Exercise \(\PageIndex{3}\).


1.14.4: Extensive and Intensive Variables
The terms, variables and properties are synonymous. Nevertheless a given thermodynamic property of a system can be classified as either intensive or extensive.
Intensive Properties. The magnitude of an intensive variable does NOT depend on the amount of chemical substance in a given closed system; e.g. density.
Extensive Properties. The magnitude of an extensive variable depends on the amount of chemical substances in a closed system; e.g. volume. Let us ask – is temperature an intensive or extensive variable? Consider two conical flasks. Flask A contains \(10 \mathrm{~cm}^{3}\) of water(\(\ell\)) at \(298 \mathrm{~K}\). Flask B contains \(5 \mathrm{~cm}^{3}\) of water(\(\ell\)) at \(298 \mathrm{~K}\). The contents of Flask are poured into Flask B.
- What is the volume of liquid in flask B? The answer is clearly \(15 \mathrm{~cm}^{3}\).
- What is the temperature of the liquid in flask B? Based on the answer to the previous question, we might answer \(596 \mathrm{~K}\), being the sum \(298 + 298\). This is clearly wrong. We have not distinguished between extensive variable variable and the intensive variable temperature. Temperature is an intensive variable as, for example, is the density of liquids.
- A quick test to decide whether a given variable is either extensive or intensive is to ask what happens to the number value if the amount of chemical substance in a system increases by a factor of two. If the variable (e.g. volume) also increases by a factor of two, the variable is extensive. If the variable ( e.g. temperature, equilibrium constant…) remains unchanged , the variable is intensive. Otherwise the variable is neither extensive nor intensive (e.g. the inverse of volume).
Footnote
[1] O. Redlich ( J. Chem.Educ.,1970, 42 ,154) presents a provocative discussion of the distinction between intensive and extensive variables.