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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}
The parameterized (polar) curve
$$r(\theta) = 2 \sin \theta$$
can be written as the parameterized (Cartesian) curve
$$(x(\theta), y(\theta)) = (2 \sin \theta \cos \theta, 2 \sin^2 \theta) = (\sin 2 \theta, \cos 2 \theta + 1);$$
since both of the components in the last expression have period $\pi$, the parameterization traces out the circular boundary twice over the interval $[0, 2 \pi]$; we only want to integrate over each direction $\theta$ once, so we should replace the interval over which we integrate $\theta$ by $[0, \pi]$.
We can see too why the triple integral you wrote evaluates to zero: The upper limit on the middle integral, is $2 \sin \theta$, and the symmetry of that function (together with the fact that $\theta$ appears nowhere else in the integral) ensures that the integral over $[\pi, 2\pi]$ has the same magnitude but the opposite sign as the desired integral over $[0, \pi]$.
Best Answer
Since the graph of $r = 2\sin \phi$ for $\phi \in [\pi,2\pi]$ overlaps with that for $\phi \in [0,\pi]$,
graph from spark notes
the upper limit for $\phi$ should be $\pi$ instead of $2 \pi$.
\begin{align} V &= \int_{0}^{\pi}\int_{0}^{2\sin\phi}\int_{0}^{2r^2}\,dz\,rdr\,d\phi \\ &= \int_{0}^{\pi}\int_{0}^{2\sin\phi}2r^3 dr\,d\phi \\ &= \int_{0}^{\pi} \frac{(2 \sin \phi)^4}{2} \,d\phi \\ &= 8\int_{0}^{\pi} \sin^4 \phi \,d\phi \\ &= 16 \int_{0}^{\pi/2} \sin^4 \phi \,d\phi \quad (\text{symmetry of sine}: \sin\left(\frac\pi2+\phi\right) = \sin\left(\frac\pi2-\phi\right)) \\ &= 16 \cdot \frac{3 \cdot 1}{4 \cdot 2} \cdot \frac{\pi}{2} \quad \text{(Wallis' formula)} \\ &= 3\pi \end{align}