Answer
$V=\frac{7\pi}{6}$
One-eighth of a hollow sphere with outer radius $2$ and inner radius $1$.
Work Step by Step
The $ \rho^{2}sin\phi d\rho d\theta d \phi$ tells us that we are dealing with spherical co-ordinates. The radius ranges from $1$ to $2$, while the angle bounds limits us to one octant, or one-eighth of the hollow sphere.
$=\int_{0}^{\pi/2}\int_{0}^{\pi/2}(\frac{\rho^{2}}{3}sin\phi)|_{1}^{2}d\phi d\theta$
$=\int_{0}^{\pi/2}\int_{0}^{\pi/2}\frac{7}{3}sin(\phi)d\phi d\theta$
$=\int_{0}^{\pi/2}(-\frac{7}{3})cos(\phi)|_{0}^{\pi/2}$
$=\int_{0}^{\pi/2}(\frac{7}{3})d \theta$
$=\frac{7\pi}{6}$
Hence, $V=\frac{7\pi}{6}$
One-eighth of a hollow sphere with outer radius $2$ and inner radius $1$.
