Answer
$\dfrac{563 \pi }{30}$
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, $V=2\pi \int_{1}^{2} (4-x)(x^3+2-4+x^2) \ dx \\ = 2\pi \int_1^2 (4-x) (x^3+x^2-2] \ dx \\= 2 \pi \int_1^2 (3x^3+4x^2-8-x^4+2x) \ dx \\= 2\pi [ \dfrac{ 3x^{4}}{4}+\dfrac{4x^{3}}{3}-8x-\dfrac{x^5}{5}+x^2]_1^2 \\=2 \pi ( \dfrac{64}{15}+\dfrac{ 307}{60}) \\= \dfrac{563 \pi }{30}$
