Answer
$\dfrac{32\pi}{5}$
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, $V=2\pi \int_{0}^{2} (x)(2-2(x^2+1)^{-2}) \ dx\\= 2\pi \int_{0}^{2} [2x-\dfrac{2x}{(x^2+1)^2}] dx \\=2 \pi [x^2+\dfrac{1}{(x^2+1)}]_{0}^{2} \\=2 \pi [2^2+\dfrac{1}{(2^2+1)} ] \\=2 \pi (4+\dfrac{1}{5} ) \\=\dfrac{32\pi}{5}$
