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\begin{align}\color{#66f}{\large V}&
=\left.\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
\dd x\,\dd y\,\dd z\right\vert_{\large{x^{2}\ +\ y^{2}\ +\ z^{2}\ <\ 9\atop
z\ >\ \pars{x^{2}\ +\ y^{2}}/8}}
=\left.\int_{0}^{2\pi}\int_{-\infty}^{\infty}\int_{0}^{\infty}r\,\dd r\,\dd z\,\dd\phi\right\vert
_{{\large r^{2}\ +\ z^{2}\ <\ 9\atop z\ >\ r^{2}/8}}
\\[5mm]&=2\pi\left.\int_{0}^{\infty}\int_{0}^{\infty}r\,\dd r\,\dd z\right\vert
_{{\large r\ <\ \root{9 - z^{2}}\atop {r\ <\ \sqrt{8z}\atop 0\ <\ z\ <\ 3}}}
=\color{#66f}{\large%
2\pi\int_{0}^{3}\int_{0}^{\min\pars{\root{9 - z^{2}},\root{8z}}}r\,\dd r\,\dd z}
\end{align}
However,
$$
\root{9 - z^{2}}<\root{8z}\ \imp\
\pars{z - 1}\pars{z + 9}>0\ \imp\ \pars{~z<-9\ \mbox{or}\ z>1~}
$$
\begin{align}
\color{#66f}{\large V}&=2\pi\int_{0}^{1}\int_{0}^{\root{8z}}r\,\dd r\,\dd z
+\int_{1}^{3}\int_{0}^{\root{9 - z^{2}}}r\,\dd r\,\dd z
\\[5mm]&=2\pi\bracks{\int_{0}^{1}\half\,\pars{8z}\,\dd z
+\int_{1}^{3}\half\,\pars{9 - z^{2}}\,\dd z}
=2\pi\braces{\left. 2z^{2}\right\vert_{0}^{1}
+\left. {9 \over 2}\,z - {1 \over 6}\,z^{3}\right\vert_{1}^{3}}
\\[5mm]&=2\pi\bracks{2 + {27 \over 2} - {27 \over 6} - {9 \over 2} + {1 \over 6}}
=\color{#66f}{\large{40 \over 3}\,\pi}
\end{align}
Best Answer
The region projected onto the $x-y$ plane gives a circle of radius $1$, and $z$ is integrated from $0$ to $4-x^2-y^2$. Therefore, using cylindrical coordinates, the integral becomes $$V = \int_0^{2\pi}\int_0^1 \int_0^{4-x^2-y^2} r\, dz\, dr\, d\theta$$