Answer
$ \dfrac{4 \pi}{3} (\sqrt {65}-\sqrt 2)$
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, $V=2\pi \int_{1}^{4} (x)(\dfrac{x}{\sqrt {1+x^3}}) \ dx$
Let us use the substitution method. So, substitute $a^2=1+x^3 \implies (2a) da=3x^2 dx$
Now, $V= 2\pi \int_{1}^{4} a^{-1} (\dfrac{2a}{3}) da \\=2 \pi \times \dfrac{2}{3}[(1+x^3)^{1/2}]_{1}^{4} \\= \dfrac{4 \pi}{3} (\sqrt {65}-\sqrt 2)$
