= Solution
Use the <spherical polar coordinates> <volume element> $dV=r^2\sin\theta\,dr\,d\theta\,d\phi$. Since $1+\cos\theta\geq0$ throughout $0\leq\theta\leq\pi$, the solid has radial range $0\leq r\leq1+\cos\theta$ and full azimuthal range $0\leq\phi<2\pi$. Its <volume> is
$$
V=\int_0^{2\pi}\int_0^\pi\int_0^{1+\cos\theta}r^2\sin\theta\,dr\,d\theta\,d\phi=\frac{2\pi}3\int_0^\pi(1+\cos\theta)^3\sin\theta\,d\theta.
$$
Put $u=1+\cos\theta$, so $du=-\sin\theta\,d\theta$ and the endpoints become $2$ and $0$. Consequently
$$
\boxed{V=\frac{2\pi}3\int_0^2u^3\,du=\frac{8\pi}3.}
$$
The 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.
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