Volume of Cylindrical Surface Bounded By Hyperboloid
\[x^2+y^2=4, \; \]
is bounded above and below by the hyperboloid \[x^2+y^2-z^2=1\]
. What is the volume of the solid enclosed?In cylindrical polar coordinates
\[x^2+y^2=r%2, \; z=z, \; \theta = \theta\]
and we can write \[r^2-z^2=1 \rightarrow 2z= \sqrt{r^2-1} \sqrt{r^2-1}\]
. The volume element is \[dV=2z r dr d \theta = 2r \sqrt{r^2-1} dr d \theta\]
. The volume is\[\begin{equation} \begin{aligned} V &= \int^{2 \pi}_0 \int^2_1 r \sqrt{r^2-1} dr d \theta \\ &= \int^{2 \pi}_0 [ \frac{2}{3} (\sqrt{r^2-1})^{\frac{3}{2}} ]^2_1 d \theta \\ &= \int^{2 \pi}_0 2 \sqrt{3} d \theta = 4 \pi \sqrt{3}\end{aligned} \end{equation}\]
