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Convert the following integral into polar coordinates :

$$\int_0^1\int_{x}^{\sqrt{2x-x^2}} (x^2 + y^2)\,dy\,dx$$

I can see that the limits of $\theta$ go from $\pi/4$ to $\pi/2$. Can someone please help me with the limits for $r$. Thanks in advance!

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Sorry I have edited it now. –  Sheetal Sarin Aug 17 '13 at 18:14
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$$\int_0^1\int_{x}^{\sqrt{2x-x^2}} (x^2 + y^2)\,dy\,dx$$

Complete the square: $$ y=2x-x^2 = -(x^2-2x) = 1-(x^2-2x+1) = 1-(x-1)^2, $$ so $$ y^2+(x-1)^2 = 1. $$ This is a circle whose center is at $(1,0)$ and whose radius is $1$.

The lower bound of integration, $y=x$, is a straight line of slope $1$ right through $(0,0)$. Draw the picture and you see that line intersecting the circle at two points: $(0,0)$ and $(1,1)$, the very leftmost point on the circle and the very highest point. And plug those coordinates into both equations and you see that this is exact. So you're looking at something that's not really a crescent but is reminiscent of a crescent.

Now draw a ray from $(0,0)$ through the first quadrant and you see it intersect that not-quite-crescent-shaped region only if the angle $\theta$ is between $\pi/4$ (half a right angle) and $\pi/2$ (straight upward). So $r$ goes from $0$ out to that curve. So you need the equation of that curve in polar coordinates.

It was $y^2+(x-1)^2=1$, but now we'll write it as $y^2+x^2-2x=0$. Then $y^2+x^2$ becomes $r^2$ and $2x$ becomes $2r\cos\theta$. We now have $r^2-2r\cos\theta=0$, so either $r=0$ or $r-2\cos\theta=0$. So the equation of that circle is $r=2\cos\theta$.

Now the integral becomes $$ \int_{\pi/4}^{\pi/2}\int_0^{2\cos\theta} r^2 \Big(r\,dr\,d\theta\Big). $$

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Thanks Michael for the lucid explanation. Its all clear to me now :) –  Sheetal Sarin Aug 19 '13 at 4:45
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If you take a ray corresponding to a value of $\theta$ between $\pi/4$ and $\pi/2$, it will begin at the origin and exit the region where $y=\sqrt{2x-x^2}$. Since squaring this equation gives $x^2+y^2=2x$ or $r^2=2r\cos\theta$ or $r=2\cos\theta$, the limits for r will be 0 and $2\cos\theta$.

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You need to convert the equation $x^2+y^2=2x$ to polar coordinates. This shouldn't be too hard. (I'm assuming you already drew a sketch of the region.)

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But there's also the problem of what the bounds of integraition should be, and the fact that $dy\,dx=r\,dr\,d\theta$. –  Michael Hardy Aug 17 '13 at 18:05
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