Answer
$\displaystyle{V=4\sqrt3\pi}$
Work Step by Step
$\displaystyle{(2)^2-x^2=1}\\ \displaystyle{x^2=3}\\ \displaystyle{x=-\sqrt3\qquad x=\sqrt3}$
$\displaystyle{A(x)=\pi\left(2\right)^2-\pi\left(\sqrt{1+x^2}\right)^2}\\ \displaystyle{A(x)=\pi\left(3-x^2\right)}$
$\displaystyle{V=\int_{-\sqrt3}^{\sqrt3}A(x)\ dx}\\ \displaystyle{V=\int_{-\sqrt3}^{\sqrt3}\pi\left(3-x^2\right)\ dx}\\ \displaystyle{V=\pi\int_{-\sqrt3}^{\sqrt3}3-x^2\ dx}\\ \displaystyle{V=2\pi\int_{0}^{\sqrt3}3-x^2\ dx}\\ \displaystyle{V=2\pi\left[3x-\frac{1}{3}x^3\right]_{0}^{\sqrt3}}\\ \displaystyle{V=2\pi\left(\left(3(\sqrt3)-\frac{1}{3}(\sqrt3)^3\right)-\left(0\right)\right)}\\ \displaystyle{V=4\sqrt3\pi}$
