<s>[INST] Find the area of the region inside: $r= 6\sin(\theta)$ but outside of $r = 1$ How do we find the area of the region inside $r = 6 \sin(\theta)$, but outside $r = 1$?
So, here's my work thus far:
First off, we know:
$r^2 = x^2 + y^2$
 and $\mathrm{sin}(\theta) = y/r$
Therefore,
$r = \sqrt{x^2+y^2}$
and $6\mathrm{\sin}(\theta) = \frac{6y}{r}$
The original equation turns into:
$\sqrt{x^2 + y^2} = 6y/r$ ; or equivalently $x^2 + y^2 = 6y$
Completing the square turns this into 
$x^2 + (y-3)^2 = 9$, a circle centered at $(0,3)$ with a radius of $3$
The area of that circle is $9\pi$, so the answer is going to be $9\pi$ - something. I'm having a hard time figuring out the something.
The other circle is $x^2 + y^2 = 1$, a circle centered at the origin with a radius of $1$.
Well, I was thinking, let's find the points of intersection by setting the two equations equal to eachother. So set $x^2 + y^2 - 1 = x^2 + y^2 - 6y$ (because they both = 0).
So cancel out the $x^2$ and $y^2$, so we get $-1 = -6y$ or $y = 1/6$.  Plug that back in and find the two x coordinates. Plugging it back into $x^2 + y^2 = 1$, so you get +/- the $\sqrt{35/36}$.  So the two intersection points are $(-\sqrt{35/36} , 1/6)$ and  $(\sqrt{35/36} , 1/6)$. So I'm thinking, alright, just integrate the following from $-\sqrt{35/36}$ to $\sqrt{35/36}$: $x^2 + y^2 - 1 - (x^2 + y^2 - 6y)dx.$ But I get the integral of $(6y - 1)dx$.  I don't understand please help!
 [/INST] You want the region where $r>1$.  That's the same as $\sin\theta > 1/6$.  The boundary of that region is where $\sin\theta=1/6$.  That happens when $\theta = \arcsin(1/6)$ and when $\theta = \frac\pi 2 - \arcsin(1/6)$.  So you want
$$
\int_{\arcsin(1/6)}^{\pi/2-\arcsin(1/6)} \frac{r^2}{2} \, d\theta = \int_{\arcsin(1/6)}^{\pi/2-\arcsin(1/6)} 18\sin^2\theta \, d\theta = \int_{\arcsin(1/6)}^{\pi/2-\arcsin(1/6)} 9 - 9\cos(2\theta)\,d\theta
$$
$$
= \left[ 9\theta - \frac92\sin(2\theta) \right]_{\arcsin(1/6)}^{\pi/2-\arcsin(1/6)}
$$
To find $\sin(2\arcsin\frac 1 6)$ use the double-angle formula for the sine.  For that, you want to remember that $\cos(\arcsin x) = \sqrt{1-x^2}$.  To find $\sin(2(\frac\pi 2 - \arcsin(1/6))) = \sin(\pi - 2\arcsin(1/6))$, remember that $\sin(\pi - \theta)$ is the same as $\sin\theta$.
And remember that $\cos(\pi-\theta) = -\cos(\theta)$.
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