Compute the triple integral over the region $x^2+y^2+z^2 \leq 2x$, $z\leq 0$ 
Compute the triple integral over the region
$D = \{x^{2} + y^{2} + z^{2} \leq 2x,\, z\leq 0\}$:
$$
\iiint_{D}\left(y^{2}z + x\right){\rm d}V
$$

I'm struggling a great deal with setting up the integration bounds. From playing around, I think the best way to go about it is using spherical coordinates as opposed to cylindrical coordinates. It's easy to make a quick sketch of the $xy$ plane to deduce that we have to be below the line $y=2x$, however then we must find the angle of intersection with the line $y=2x$ with the circle $x^2+y^2=1$, and using polar coordinates, we can find that it corresponds to $\arctan2$ and then of course $\arctan2 + \pi$, and by $z \leq 0$, we ought to obtain that $\frac{\pi}{2}\leq\phi \leq \pi$. I'm unsure about the $R$ bound (and everything previously stated to be honest), but using the provided inequality, we would obtain that $R \leq2\cos\theta \sin\phi$, but by my sketch it would make more sense to have $0 \leq R \leq1$.
Thus, my suggested integral would look something like:
$$\int_{\arctan{2}+\pi}^{\arctan{2}}d\theta\int_{\frac{\pi}{2}}^{\pi}d\phi\int_{0}^1(R^2 \sin{\phi})(R^3\sin^2{\theta}\sin^2{\phi}\cos{\phi+R\cos{\theta}\sin{\phi}})dR$$
which is obviously a nightmare. Any help/ hints on how to establish the bounds would be very appreciated!
 A: First, let $x\to x+1$ to simplify the integral
$$I=\iiint_{D}y^2z+x \ dV= \iiint_{D’}y^2z+x+1 \ dV= \frac{2\pi}3+\iiint_{D’}y^2z\ dV
$$
where $D’= \{$ $x^2+y^2+z^2 \leq 1$, $z\leq 0\}$, $\iiint_{D’}x \ dV= 0$ due to symmetry and $\iiint_{D’} dV=\frac{2\pi}3 $ is the half-sphere volume. Then, evaluate the remaining integral in spherical coordinates
\begin{align}
\iiint_{D’}y^2z\ dV=& \
\frac12\iiint_{D’}(x^2+y^2)z \ dV \\
=&\ \frac12\int_{\pi/2}^\pi\int_0^{2\pi}\int_0^1 \cos\phi\sin^3\phi \ r^5 \ dr d\theta d\phi = -\frac\pi{24}
\end{align}
Thus, $I= \frac{2\pi}3-\frac\pi{24}=\frac{5\pi}{8}$.
A: You should recognize — complete the square — that $x^2+y^2+z^2=2x$ is a sphere of radius $1$ centered at $(1,0,0)$. To make life more convenient so that I can easily use spherical coordinates, I am going to switch the $xyz$-axes to the $zxy$-axes. Thus, the problem becomes this:
Integrate $x^2y+z$ over the region $D' = \{x^2+y^2+z^2\le 2z, y\le 0\}$. In spherical coordinates, the sphere is given by $\rho = 2\cos\phi$, $0\le\phi\le\pi/2$. (You can see this algebraically, by writing $\rho^2 = 2\rho\cos\phi$, or by basic right triangle geometry.) So we have
$$\int_{-\pi}^0\int_0^{\pi/2}\int_0^{2\cos\phi} (\rho^3\sin^3\phi\cos^2\theta\sin\theta + \rho\cos\phi)\rho^2\sin\phi\,d\rho\,d\phi\,d\theta.$$
Note that $\int_{-\pi}^0 \cos^2\theta\sin\theta\,d\theta = -2/3$ and so the first term integrates to
$$-\frac23\int_0^{\pi/2}\int_0^{2\cos\phi} \rho^5\sin^4\phi\,d\rho\,d\phi = -\frac{\pi}{24}.$$
The second term gives
$$\pi\int_0^{\pi/2}\int_0^{2\cos\phi}  \rho^3\cos\phi\sin\phi\,d\rho\,d\phi = \pi\int_0^{\pi/2}4\cos^5\phi\sin\phi\,d\phi = \frac{2\pi}3.$$
Thus, the final answer is $\boxed{\pi\left(\frac23-\frac1{24}\right) = \frac{5\pi}8}$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Hereafter, $\ds{\bracks{\cdots}}$ is an $\ds{Iverson's\ Bracket}$.
With $\ds{D \equiv \braces{\pars{x,y,z} \mid x^{2} + y^{2} + z^{2} \leq 2x\ \wedge\  z\leq 0}}$,
\begin{align}
& \color{#44f}{\large\iiint_{D}\pars{y^{2}z + x}\dd x\,\dd y\,\dd z} =
\int_{-\infty}^{0}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\bracks{x^{2} + y^{2} + z^{2} \leq 2x}\pars{y^{2}z + x}
\dd x\,\dd y\,\dd z
\\[5mm] = & \
\int_{0}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\bracks{\pars{x - 1}^{2} + y^{2} + z^{2} \leq 1}
\pars{-y^{2}z + x}\dd x\,\dd y\,\dd z
\\[5mm] = & \
\int_{0}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\bracks{x ^{2} + y^{2} + z^{2} \leq 1}
\pars{-y^{2}z + x + 1}\dd x\,\dd y\,\dd z
\end{align}
In using Spherical Coordinates:
\begin{align}
& \color{#44f}{\large\iiint_{D}\pars{y^{2}z + x}\dd x\,\dd y\,\dd z}
\\[5mm] = & \
\int_{0}^{2\pi}\int_{0}^{\pi/2}\int_{0}^{1}
\braces{\vphantom{A^{A^{A}}}-\braces{r\sin\pars{\theta}\sin\pars{\phi}}^{2}\braces{r\cos\pars{\theta}} +
r\sin\pars{\theta}\cos\pars{\phi} + 1}\ \times
\\ & \ \phantom{\int_{0}^{2\pi}\int_{0}^{\pi/2}\int_{0}^{1}\,\,\,}
r^{2}\sin\pars{\theta}\,\dd r\,\dd\theta\,\dd\phi
\\[5mm] = & \
\int_{0}^{\pi/2}\int_{0}^{1}
\braces{\vphantom{A^{A^{A}}}-r^{3}\sin^{2}\pars{\theta}\cos\pars{\theta}\
\overbrace{\int_{0}^{2\pi}\sin^{2}\pars{\phi}\dd\phi}^{\ds{= \pi}}\ +\
r\sin\pars{\theta}\ \overbrace{\int_{0}^{2\pi}\cos\pars{\phi}\dd\phi}^{\ds{= 0}} + \int_{0}^{2\pi}\dd\phi\ }\ \times
\\ & \ \phantom{\int_{0}^{2\pi}\int_{0}^{\pi/2}\int}
r^{2}\sin\pars{\theta}\,\dd r\,\dd\theta
\\[5mm] = & \
-\pi\pars{\int_{0}^{1}r^{5}\dd r}
\braces{\int_{0}^{\pi/2}\sin^{3}\pars{\theta}\cos\pars{\theta}\,\dd\theta} +
2\pi\pars{\int_{0}^{1}r^{2}\dd r}
\braces{\int_{0}^{\pi/2}\sin\pars{\theta}\,\dd\theta}
\\[5mm] = & \ \bbx{\color{#44f}{\large{5\pi \over 8}}} \approx 1.9635 \\ &
\end{align}
