Answer
$\dfrac{425 \pi }{3}$
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}=1+\dfrac{3x}{\pi} $ and $ R_{inside}=\sec x$
Now, $V=\pi \int_{-65}^{0} [9-(\sqrt [4] {16-y})^2)] \ dy \\ = \pi [9y+\dfrac{2(16-y)^{3/2}}{3}]_{-65}^0 \\=\pi [\dfrac{128}{3}-(-585+\dfrac{2(9^3)}{3} \\=\dfrac{425 \pi }{3}$
