Answer
$\dfrac{16 \pi}{3}$
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, $V=2\pi \int_{1}^{3} (4-x) (-x^2+4x-3) \ dx \\= 2\pi \int_{1}^{3} (-8x^2+19x +x^3-12) \ dx \\=2 \pi \times [\dfrac{19x^2}{2}+\dfrac{x^4}{4}-12x-\dfrac{8x^3}{3}]_1^3 \\=\dfrac{16 \pi}{3}$
