Integral using spherical coordinates I am trying to compute the volume of the following set :
intersection of cylinder $x^2 + y^2 \leq R$  and sphere $x^2 + y^2 + z^2 \leq 4R^2$.
I am having trouble setting up the integral properly after transforming to spherical coordinates I am not sure where the sphere and the cylinder meet and how to compute the volume of that top part.
I could use some help. Thank you 
 A: Actually in this case you don't need neither spherical coordinates nor cilindrical. You can find volume using double integral.
If $\sqrt{R}\le 2R$ it is equal
$$
V=2\int\int_{x^2+y^2\le R}\sqrt{4R^2-x^2-y^2}dxdy=2\int_{0}^{2\pi}\left(\int_{0}^{\sqrt R}r\sqrt{4R^2-r^2}dr\right)d\varphi
.$$
I  think you can manage to compute last integral.
A: Actually using spherical coordinates:  let $\alpha=\arcsin(1/(2\sqrt{R}))$ (for use in a couple of the limits -- this is the $\phi$ angle where the sphere and cylinder intersect, integral with respect to $\phi$ is split into two parts there), and computing the volume of the upper half and doubling, the total volume is
$$
2\left[
\int_0^{2\pi}\int_0^\alpha\int_0^{2R} \rho^2\sin\phi 
d\rho d\phi d\theta \\
+\int_0^{2\pi}\int_\alpha^{\pi/2}\int_0^{\sqrt{R}\csc\phi}
\rho^2\sin\phi d\rho d\phi d\theta
\right]
$$
OK -- now are you absolutely certain this is how you want to do this?  I agree with others who have suggested that cylindrical coordinates may be easier.
