Answer
$\dfrac{1400 \pi}{3}$
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, $R_{outside}=16-2x$ and $ R_{inside}=6$
Now, $V=\pi \int_0^5 [(16-2x)^2 -(6)^2] \ dx \\ = \pi \int_0^5 [256+4x^2-64x-36] \ dx \\=\pi \int_0^5 [220+4x^2-64 x] \ dx \\=\pi (220+\dfrac{4x^3}{3}-32x^2]_0^5 \ dx \\= \dfrac{1400 \pi}{3}$
