I want to find $\int_0^2 \int_ {\sqrt{2x-x^2}}^{\sqrt{4-x^2}} \sqrt{4-x^2-y^2}dydx$.
My idea is to do a transformation to polar coordinates;
$\iint \sqrt{4-r^2} rdrd\theta$, but I'm unsure about the bounds. Plotting the bounds in $f(x,y)$ I realize I'm dealing with the area between a smaller circle centered at (1, 0) and r=1 and a bigger circle centered at (0,0) with r = 2. (Under the ball $\sqrt{4-x^2-y^2}$ Any ideas? Thank you