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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.

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  • $\begingroup$ Rather in the first integral the superior limit is $\sqrt{2\sin \theta}$. $\endgroup$ – Valerin Dec 11 '13 at 16:19
  • $\begingroup$ Even then, it is still wrong as per Wolfram Alpha. $\endgroup$ – user1251385 Dec 11 '13 at 16:21
  • $\begingroup$ The better way to change the coordinates is to draw the region of integration. If you do this is less probable that you be wrong. $\endgroup$ – Valerin Dec 11 '13 at 16:27
  • $\begingroup$ Yes as @John say the upper limit in the second integral is $\pi/2$ instead $\pi$. +1 $\endgroup$ – Valerin Dec 11 '13 at 16:30
  • $\begingroup$ Since the y-axis is bounded by 1, it doesnt cover the whole circle, so u have to find the angle at 1. $\endgroup$ – John Dec 11 '13 at 16:32
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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:

http://i.stack.imgur.com/vPDCi.png

We know the circle is of origin (0,1)

$\tan(\theta)=1 \implies \theta=\frac{\pi}{4}$

So, $$\int_0^{\pi/4} \int_0^{2\sin\theta}r(1-r^2)drd\theta$$

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  • $\begingroup$ say if the bounds of integration were from $-4 \le x \le 4$ and $0 \le y \le \sqrt{16 - y^2}$ $\endgroup$ – user1618033988749895 Dec 11 '20 at 4:54
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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):

enter image description here

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  • $\begingroup$ say if the bounds of integration were from $-4 \le x \le 4$ and $0 \le y \le \sqrt{16 - y^2}$ $\endgroup$ – user1618033988749895 Dec 11 '20 at 4:54
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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 $\pi/4$.

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