Answer
$\dfrac{2 \pi}{3}[(1+c^2)^{3/2}-1] $
Work Step by Step
The shell method to compute the volume of a region: The volume of a solid obtained by rotating the region under $y=f(x)$ over an interval $[m,n]$ about the y-axis is given by:
$V=2 \pi \int_{m}^{n} (Radius) \times (height \ of \ the \ shell) \ dy=2 \pi \int_{m}^{n} (y) \times f(y) \ dy$
Now, $V=2\pi \int_{0}^{c} (x) (\sqrt {1+x^2}) \ dx$
Consider $a=1+x^2 \implies dx=\dfrac{da}{2x}$
Now, $V= 2\pi \int_{0}^{c} (x) (\sqrt {a})\times \dfrac{1}{2x} \ da \\=2 \pi [\dfrac{1}{3} a^{3/2}] \\=\dfrac{2 \pi}{3} (1+x^2)^{3/2} \\=\dfrac{2 \pi}{3} [(1+x^2)^{3/2}]_0^c \\=\dfrac{2 \pi}{3}[(1+c^2)^{3/2}-1] $
