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!