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$$\int_0^1\int_0^\sqrt{2y-y^2}(1-x^2-y^2)dxdy$$

I tried to transform it to polar form, but the problem is to find the limits of integration.

$0\le x\le \sqrt{2y-y^2}$ is the right half of the circle $x^2+(y-1)^2=1$ , then $r=2sin(\theta)$. And $0\le y\le 1$ makes the area is the lower-right quarter of the circle.

enter image description here What are the limits of $r$ and $\theta$ ?

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I'm not sure I understand the question correctly, since it seems to me you've already solved it yourself – the part that you may be missing is that the polar coordinates in this case shouldn't be with respect to the origin but with respect to the centre of the circle:

\begin{align} x&=r\cos\theta\;,\\ y&=r\sin\theta+1\;. \end{align}

Then the limits of $r$ and $\theta$ are evident from what you wrote.

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    $\begingroup$ Aha, got it. I should first translate the coordinate axes to the center of the circle. Problem solved. Thanks $\endgroup$ – Mohamed Mostafa May 30 '16 at 11:16

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