Finding volume triple integral

Question

Find the volume of the solid enclosed on the outside of the sphere $r=2$ and on the inside by the surface $r=1+\cos\phi$.

I am not sure how to set it up.

Answer 1

Since $d^3\mathbf{x}=r^2dr\sin\theta d\theta d\phi$, your integral is $$V:=\int_0^\pi\sin\theta d\theta \int_0^{2\pi}d\phi\int_{1+\cos\phi}^2 r^2 dr.$$ Since odd powers of $\cos\phi$ will integrate to $0$, $$V=\frac{2}{3}\int_0^{2\pi}(7-3\cos^2\phi)d\phi=\frac{22\pi }{3}.$$