Answer
$32 \pi$
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:
$V=\pi \int_{m}^{n} (R^2_{outside}-R^2_{inside}) \ dy$
Now, $V=\pi \int_{0}^{4} [(8-x)^2-(8-2 \sqrt x)^2] \ dy \\ = \pi \int_{0}^{4} [(64-16x+x^2-(64-32x^{1/2}+4x) \ dx \\=\pi [-10x^2+\dfrac{x^3}{3}+\dfrac{32 x^{3/2}}{3/2}]_0^4 \\= 32 \pi$
