Answer
$2 \pi (\sqrt 5-1)$
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:
$V=2 \pi \int_{m}^{n} (Radius) \times (height \ of \ the \ shell) \ dy=2 \pi \int_{m}^{n} (x) \times f(x) \ dx$
Now, $V=2\pi \int_{0}^{2} (x) (\dfrac{1}{\sqrt {x^2+1}}) \ dx\\= 2\pi \int_{0}^{2} (\dfrac{x}{\sqrt {x^2+1}}) \ dx \\=2 \pi [\dfrac{-4}{3x^3} +\dfrac{1}{2x^2}]_{-3}^{-1} \\= 2\pi [(1+x^2)^{1/2}]+0^2 \\=2 \pi (\sqrt 5-1)$
