[tex]\bf x^2+(y-4)^2=16\implies x^2=16-(y-4)^2
\\\\\\
x=\sqrt{16-(y-4)^2}\implies x=\sqrt{16-(y^2-8y+16)}
\\\\\\
x=\sqrt{8y-y^2}\\\\
-------------------------------\\\\
0=\sqrt{8y-y^2}\implies 0=8y-y^2\implies 0=y(8-y)
\implies
y=
\begin{cases}
0\\
8
\end{cases}\\\\
-------------------------------\\\\
\displaystyle \stackrel{\textit{disk method}}{\int\limits_{0}^8~\pi (\sqrt{8y-y^2})^2\cdot dy}\implies \pi \int\limits_{0}^8~ 8y-y^2\cdot dy[/tex]
[tex]\bf \displaystyle \pi \int\limits_{0}^8~8y\cdot dy - \pi \int\limits_{0}^8~y^2\cdot dy\implies \pi \left[ 4y^2~-~\cfrac{y^3}{3} \right]_{0}^8
\\\\\\
\pi \left[ \left[ 256-\cfrac{51}{3} \right]~-~[0] \right]\implies \pi \cdot \cfrac{256}{3}\implies \cfrac{256\pi }{3}\implies 85\frac{\pi }{3}[/tex]
