Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2001 / Paper 2 / 8 / b / Solution

Use the spherical polar coordinates volume element $dV=r^2\sin \theta drd\theta d\varphi$. Since $1+\cos \theta \ge 0$ throughout $0\le \theta \le \pi$, the solid has radial range $0\le r\le 1+\cos \theta$ and full azimuthal range $0\le \varphi <2\pi$. Its volume isPut $u=1+\cos \theta$, so $du=-\sin \theta d\theta$ and the endpoints become $2$ and $0$. ConsequentlyThe polar equation describes the entire three-dimensional surface, so the spherical polar coordinates volume element, rather than a two-dimensional polar area element, is required.