Integral using spherical coordinates

Question

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

Answer 1

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.

Answer 2

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.