Evaluate the given integral by changing to polar coordinates.

$\displaystyle \int_{0}^{2}\,\,\,\,\,\,\,\,\,\,\int_{0}^{\sqrt{2x-x^{2}}}\sqrt{4-x^{2}-y^{2}}dxdy$

Thanks in advance.

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- Mar 21st 2009, 08:47 AMBiscaima double integral
Evaluate the given integral by changing to polar coordinates.

$\displaystyle \int_{0}^{2}\,\,\,\,\,\,\,\,\,\,\int_{0}^{\sqrt{2x-x^{2}}}\sqrt{4-x^{2}-y^{2}}dxdy$

Thanks in advance. - Mar 21st 2009, 09:46 AMOpalg
The integral is $\displaystyle \int_0^2\int_0^{\sqrt{2x-x^2}}\!\!\!\sqrt{4-x^{2}-y^{2}}\,dy\,dx$. You need to work out the region over which the integral takes place, and for that you must

**draw a picture**.

Notice that y goes from 0 to $\displaystyle \sqrt{2x-x^2}$. The upper limit of that interval occurs when $\displaystyle y = \sqrt{2x-x^2}$. Square both sides of that equation, rearrange it a bit, and you see that it can be written as $\displaystyle (x-1)^2+y^2=1$. You will recognise that as the equation of the circle centred at (1,0) with radius 1. But the lower limit for y is 0, so the region of integration is the upper half of that circle.

That is stage 1 of the solution. Stage 2 is to describe that region in terms of polar coordinates.**Have you drawn a picture of that semicircular region yet?**If so, you should be able to see that $\displaystyle \theta$ goes from 0 to $\displaystyle \pi/2$ and, for each fixed value of $\displaystyle \theta$, r goes from 0 to $\displaystyle 2\cos\theta$.

Once you have got that far, you can write the integral as $\displaystyle \int_0^{\pi/2}\int_0^{2\cos\theta}\!\!\sqrt{4-r^2}\;rdr\,d\theta$, and the rest should be easy.