Evaluate the iterated integral by converting to polar coordinates:
$\large \int^2_0 \int^{\sqrt{2x-x^2}}_0 xy~dy~dx$
I successfully completed most of the problem; however, I had difficulty with the limits of integration for the variable r.
To properly describe the region with polar coordinates, I drew the graph; from this, I see that r takes on the values from 0 to the points on the graph $y= \sqrt{2x-x^2}$, including the values inside. In polar form, the function would be $r^2-2r\cos\theta=0$. Unsure of how to proceed, I looked at the answer key. I saw that the upper limit of integration for r was $2\cos\theta$. From this, suspected that they divided by r. Isn't this technically a fallacious way of proceeding, though? Doesn't r take on the value 0?
The region of integration is a semicircle. It is the top half of a circle of radius $1$ centered at $(1,0)$. If I wanted to do this in polar coordinates, I would move the origin to the center of the circle by defining $u=x-1, du=dx$ Now I have $$\int_0^2\int_0^{\sqrt{2x-x^2}}xy\; dy\ dx=\int_0^2\int_0^{\sqrt{1-u^2}}(u+1)y\; dy \; du$$. Now the integral is over the top half of the unit circle centered at the origin, so $r$ goes from $0$ to $1$ and $\theta$ goes from $0$ to $\pi$
Without the shift, $\theta$ would range from $0$ to $\frac \pi 2$ and $r$ would range from $0$ to $2 \cos \theta$. You are correct that they divided by $r$ to get the upper limit, however $r$ is not zero at the upper limit (except at $\theta=\frac \pi 2$ where you should strictly make a limit argument). $r$ does go to zero in the region of integration, but not at the edge where the division was done.