Answer
$\dfrac{ 38 \pi}{5}$
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, $V=2\pi \int_{0}^{1} y (19-y^3-6y^2-12y) \ dy\\= 2 \pi [ \dfrac{19y^2}{2}-\dfrac{y^{5}}{5}-\dfrac{3y^4}{2}-4y^3]_0^1 \\= 2\pi [\dfrac{19}{2}-\dfrac{1}{5}-\dfrac{3}{2}-4-0] \\=\dfrac{ 38 \pi}{5}$
