Answer
$40 \pi$
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:
$V=2 \pi \int_{m}^{n} (Radius) \times (height \ of \ the \ shell) \ dy=2 \pi \int_{m}^{n} (x) \times f(x) \ dx$
Now, $V=2\pi \int_{0}^{2} (x+3) (4-x^2) \ dx \\ = 2\pi \int_0^2 [4x-x^3+12-3x^2]_0^2 \\=2\pi [2x^2-\dfrac{x^4}{4}+12x-x^3]_0^2 \\=40 \pi$
