Answer
$$\frac{4802\pi }{5}$$
Work Step by Step
Given $$x=4-y,\ \ x=16-y^2 $$
We first find the intersection points
\begin{align*}
4-y&=16-y^2\\
y^2-y -12&=0
\end{align*}
Then $y=-3,\ 4$
Then the volume of the solid obtained by rotating the region enclosed by the graph of
$x=4-y,\ \ x=16-y^2$ about the $y-$axis is given by
\begin{align*}
V&=\pi\int_{-3}^{4}[f(y)]^2-[g(y)]^2dy\\
&=\pi \int_{-3}^{4}\left[\left(16-y^{2}\right)^{2}-(4-y)^{2}\right] d y\\
&=\pi \int_{-3}^{4}\left[y^{4}-33 y^{2}+8 y+240\right] d y\\
&=\pi \left(\frac{y^{5}}{5}-\frac{33 y^{3}}{3}+\frac{8 y^{2}}{2}+240 y\right)\bigg|_{-3}^{4}\\
&=\frac{4802\pi }{5}
\end{align*}
We use the Mathematica program to plot the solid.
