Answer
$4 \pi \sqrt 3$
Work Step by Step
The Washer method to compute the volume of revolution: When the function $f(x)$ is continuous and $f(x) \geq g(x) \geq 0$ on the interval $[m,n]$, then the volume of a solid obtained by rotating the region under $y=f(x)$ over an interval $[m,n]$ about the x-axis is given by:
$V=\pi \int_{m}^{n} (R^2_{outside}-R^2_{inside}) \ dy$
Now, $V=\pi \int_{-\sqrt 3}^{\sqrt 3} [(\sqrt {4-y^2})^2-1] \ dy \\ = \pi \int_{-\sqrt 3}^{\sqrt 3} (3-y^2) \ dy \\=\pi (3y-\dfrac{y^3}{3}]t_{-\sqrt 3}^{\sqrt 3}\\=\pi (6 \sqrt 3-2 \sqrt 3) \\= 4 \pi \sqrt 3$
