<s>[INST] Finding a volume using the method of cylindrical shells which is generated by a parabola This problem is from the 7th edition of the book "Calculus and Analytic Geometry" by George Thomas
and Ross Finney. It is problem number 3 of section 5.4
Problem:
Find the volume generated when the region bounded by the given curves and lines is revolved about the
x-axis. (Note: $x = 0$ is the y-axis and $y = 0$ is the x-axis.)
\begin{align*}
y &= 3x - x^2 \\
y &= x
\end{align*}
Answer:
I am going to use the method of cylindrical shells to compute the volume. Let $V$ be the volume we are asked to find.
The general form for $V$ when we are revolving about the x-axis is:
$$ V = \int_a^b 2 \pi x f(x) \,\, dx $$
where $2 \pi x$ represents the circumference of the region. Since we are going around the x-axis $a = 0$. To find $b$ be we set up the following equation:
\begin{align*}
3x - x^2 &= x \\
2x - x^2 &= 0 \\
x^2 - 2x &= 0 \\
x(x-2) &= 0 \\
x = 0 \,&\text{ or } \, x = 2 \\
V &= \int_0^2 2 \pi x (  3x - x^2 - x) \,\, dx \\
V &= 2 \pi \int_0^2 x (  2x - x^2 ) \,\, dx \\
\int_0^2 x (  2x - x^2 ) \,\, dx &= \int_0^2 2x^2 - x^3 \,\, dx \\
\int_0^2 x (  2x - x^2 ) \,\, dx &= \frac{2x^3}{3} - \frac{x^4}{4} \Big|_0^2 \\
\int_0^2 x (  2x - x^2 ) \,\, dx &=  \frac{2(8)}{3} - \frac{16}{4} = \frac{16}{3} - 4 \\
\int_0^2 x (  2x - x^2 ) \,\, dx &= \frac{4}{3} \\
%
V &=  2 \pi \left(   \frac{4}{3} \right)  \\
V &= \frac{8 \pi}{3}
\end{align*}
The book's answer is:  $\frac{56\pi}{15}$
Where did I go wrong?
 [/INST] You said that you would use shells but you have actually used washers. The volume should be $$V=\int_0^2 2\pi y f(x) dx$$ Here $f(x)=(3x-x^2)-x$ is the length of the rotated vertical element and $y=\frac12 ((3x-x^2)+x)$ is the distance of the centroid (= mid-point) of the rotated element from the $x$ axis.
If you were to use shells then the volume would be $$V=\int_0^{9/4} 2\pi y g(y) dy$$ This is more difficult. The upper limit of integration is the peak of the parabola. $y$ is the distance of the rotated horizontal element from the $x$ axis. $g(y)$ is the length of the rotated horizontal element. It is a compound function. For $y=0..2$ it spans between the left side of the parabola and the straight line : $g(y)=x_0-x_1$ where $x_0=y$ and $x_1$ is the smaller solution of $x^2-3x+y=0$. For $y=2..9/4$ it spans between the left and right sides of the parabola : $g(y)=x_2-x_1$ where $x_2$ is the larger solution of $x^2-3x+y=0$.
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