#### Answer

$c \pi$

#### 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:
$V=\pi \int_{m}^{n} (R^2_{outside}-R^2_{inside}) \ dy$
where, $R_{outside}=\sqrt {1+x^2}$ and $ R_{inside}=x$
Now, $V=\pi \int_0^c [(\sqrt {1+x^2})^2 -(x)^2] \ dx \\ = \pi \int_0^c (1+x^2-x^2) \ dx \\=\pi [ x]_0^c \ dx \\=\pi (c-0) \\=c \pi$