Answer
$16 \pi$
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, $V=2\pi \int_{0}^{2} x (x^2+1) \ dx\\= 2 \pi \int_0^1 (x^3+x) \ dx \\= 2\pi [\dfrac{x^4}{4}+\dfrac{x^2}{2}]_0^2 \\=2 \pi [\dfrac{16}{4}+\dfrac{4}{2}-0] \\=16 \pi$
