Answer
$\dfrac{ 32 \pi }{3}$
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}=(x-4) $ and $ R_{inside}=2 \sqrt x-4$
Now, $V=\pi \int_{0}^{4} [(x-4)^2-(2\sqrt x-4)^2] \ dx \\ = \pi \int_0^4 [(x^2-8x+16) -(4x-16 \sqrt x+16) ] \ dx \\=\pi \int_0^4 [x^2-12x+16 x^{1/2} ] \ dx \\=\dfrac{ 32 \pi }{3}$
