Convert the integral from rectangular to cylindrical coordinates and solve
I think I know how to do this, but I just want to double check my method. So assuming I have the below problem:
$$\int^2_0\int^\sqrt{2x-x^2}_0xy dy dx$$
Since:
$$x=rcos\theta$$
$$y=rsin\theta$$
Is it then true that the integral becomes:
$$\int^2_0\int^\sqrt{2rcos\theta-r^2cos^2\theta}_0rcos\theta\; rsin\theta\; r\;dr\; d\theta$$
$$\int^2_0\int^\sqrt{2rcos\theta-r^2cos^2\theta}_0r^3cos\theta\; sin\theta\; dr\; d\theta$$
$$\int^2_0 \frac{r^4}{4}cos\theta\; sin\theta\; |^{\sqrt{2rcos\theta-r^2cos^2\theta}}_0\;d\theta$$
$$\frac{1}{4}\int^{2}_{0}(r\;cos\theta\; (2-r\;cos\theta))^4cos\theta\; sin\theta\; d\theta$$
Is this correct thus far?
EDIT
Attempting to convert to cylindrical coordinates again:
$$y=\sqrt{2x-x^2}$$
$$r\;sin\theta\;=\sqrt{2r\;cos\theta-r^2cos^2\theta}$$
$$r^2\;sin^2\theta=2r\;cos\theta-r^2cos^2\theta$$
$$r^2sin^2\theta-2r\;cos\theta+r^2cos^2\theta=0$$
$$r^2(sin^2\theta+cos^2\theta)=2r\;cos\theta$$
$$r=2cos\theta$$
$$\int^2_0\int^{2cos\theta}_0r^3\;cos\theta\;sin\theta\;dr\;d\theta$$
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$\begingroup$I think the first limits are correct, but not the second, the angle would be only between 0 and pi/2 based on the area if you draw on a graph.
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