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Rewrite the iterated integral $$\int_0^1 \int_0^{\sqrt{2y - y^2}} (1 - x^2 - y^2)\,dx\,dy$$ in polar coordinate form. Do not evaluate the integral.
Here is my answer:
$$\int_0^\pi\int_0^{2sin\theta}(1-r^2)rdrd\theta$$
I evaluated both double integrals using Wolfram Alpha and it seems my answer is wrong.
From the iterate integral, we know that the region is $x=\sqrt{2y-y^2}$ with some rearranging one can get $x^2+y^2-2y+1-1=0$ which is $x^2+(y-1)^2=1$. Equivalently in polar it translates to $r^2=2r\sin(\theta) \iff r=2\sin\theta$. The iterated integral also has a constriction on y: $0<y<1$. From the figure below, its obvious that this doesn't cover the whole circle. Therefore, we have to find the angle:
We know the circle is of origin (0,1)
$\tan(\theta)=1 \implies \theta=\frac{\pi}{4}$
So, $$\int_0^{\pi/4} \int_0^{2\sin\theta}r(1-r^2)drd\theta$$
I decided to randomly check my textbook and by coincidence, it turns out this problem was taken from there.
Here is the correct solution (I checked it with Wolfram Alpha):
The inner bound should be $2 sin \theta$. You were right the first time. This is because when $x = \sqrt{2y-y^2}$, $r = 2sin \theta$. So that gives the upper limit when you integrate with respect to $r$. You are integrating over part of a disk of radius 1 centered at (0,1) (You figure this out by doing some algebra with $x^2 + y^2 = 2y$, namely completing the square after subtracting $2y$ on both sides.) To figure out the part, notice that in the original integral $x$ goes from 0 to $\sqrt{2y-y^2}$, so you're only integrating over positive $x$ values. So you're integrating over part of the right half of the disk. The fact that $y$ is going from 0 to 1 tells you that you're integrating over only the bottom half of the disk. This tells you that $\theta$ is going from 0 to $\pi/4$.
