Answer
$\dfrac{1872 \pi}{5}$
Work Step by Step
The Washer method to compute the volume of revolution: When the function $f(x)$ is continuous and $f(x) \geq g(x) \geq 0$ on the interval $[m,n]$, then the volume of a solid obtained by rotating the region under $y=f(x)$ over an interval $[m,n]$ about the x-axis is given by:
$V=\pi \int_{m}^{n} (R^2_{outside}-R^2_{inside}) \ dy$
where, $R_{outside}=14-x$ and $ R_{inside}=x^2+2$
Now, $V=\pi \int_0^3 [(14-x)^2 -(x^2+2)^2] \ dx \\ = \pi \int_0^3 [196+x^2-28x -x^4-4-4x^2] \ dx \\=\pi \int_0^3 [-x^4-3x^2-28 x +192 ] \ dx \\=\pi (-\dfrac{x^5}{5}-x^3-14x^2+192 x)_0^3 \ dx \\=\dfrac{1872 \pi}{5}$
