#### Answer

$2 \pi (c+\dfrac{c^3 }{3}) $

#### Work Step by Step

The disk method to compute the volume of revolution: When the function $f(x)$ is continuous and $f(x) \geq 0$ 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:
$V= \pi \int_{m}^{n} (R^2) \ dx$
Now, $V=2\pi \int_{-c}^{c} (\sqrt {1+x^2})^2 \ dx \\ =2\pi [x+\dfrac{1}{3}x^3]_{-c}^{c} \\=\pi [c+\dfrac{c^3}{3}-(-c+\dfrac{(-c^3)}{3}] \\=\pi [c+\dfrac{c^3}{3}+c +\dfrac{c^3}{3}]\\=2 \pi (c+\dfrac{c^3 }{3}) $