Answer
$56 \pi$
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, $V=2\pi \int_{0}^{2} (x+3) (x^2+2) \ dx\\= 2 \pi \int_0^2 (x^3+2x+3x^2+6) \ dx \\= 2\pi [\dfrac{x^4}{4}+x^2+x^3+6x]_0^2 \\=2 \pi [\dfrac{16}{4}+4+8+12-0] \\=56 \pi$
