Answer
$\displaystyle{V=\frac{4\pi}{3}}$
Work Step by Step
$\displaystyle{A(y)=\pi\left(\sqrt{y^2-1}\right)^2}\\
\displaystyle{A(y)=\pi\left(y^2-1\right)}$
$\displaystyle{V=\int_{1}^{2}A(x)\ dy}\\
\displaystyle{V=\int_{1}^{2}\pi\left(y^2-1\right)\ dy}\\
\displaystyle{V=\pi\int_{1}^{2}y^2-1\ dy}\\
\displaystyle{V=\pi\left[\frac{1}{3}y^3-y\right]_{1}^{2}}\\
\displaystyle{V=\pi\left(\left(\frac{1}{3}(2)^3-(2)\right)-\left(\frac{1}{3}(1)^3-(1)\right)\right)}\\
\displaystyle{V=\frac{4\pi}{3}}$
