Evaluating the bounds for a triple integral I've working on the problem:
Evaluate $\iiint_Q$ $1/(x^2 + y^2 + z^2)$, where Q is the solid region ABOVE the xy-plane (and we must do this in spherical coordinates).
What I've done thus far is noted that the denominator inside the triple integral sign ($x^2+y^2+z^2)$ is equal to $\rho^2$. Thus, it becomes 1/$\rho^2$.
For $\phi$, it will be between 0 and $\pi/2$ (since it can't go the full 180 degrees  due to the bound in the xy-plane.
For $\theta$, I have said that it uses the normal bounds between 0 and 2$\pi$.
However I see no way to find the value for $\rho$? We were given no hints about it whatsoever. So any help on how I would go about this is greatly appreciated!
Thank you!
 A: Using $0$ and infinity for the bounds for $\rho$, your integral becomes improper:
$\begin{align} \int_0^\infty \int_0^{\pi/2}\int_0^{2\pi} \frac{1}{\rho^2}\rho^2\sin\phi \; d\theta \ d\phi \ d\rho &=\int_0^\infty \int_0^{\pi/2}\int_0^{2\pi} \sin\phi \; d\theta \ d\phi \ d\rho \\
&=\lim_{a\to \infty} \int_0^a \int_0^{\pi/2}\int_0^{2\pi} \sin\phi \; d\theta \ d\phi \ d\rho
\end{align}$
You can integrate this as usual and then evaluate the limit.
A: $\renewcommand{\r}{\rho}$ 
Since you have to reach the entire upper half-space of $\mathbb{R}^{3}$, you need to let $\rho$ to vary from $0$ to $+\infty$. 
As you have already noted, the cutting out lower half-plane happens when you restrict your polar (aka elevation) angle $\phi \in [0, \pi/2]$. 
Then we have
$$
\left.
\begin{aligned}
&\begin{cases}
x & = \r \sin \phi \cos \theta,\\
y & = \r \sin \phi \sin \theta,  \\
z &= \r \cos \phi,
\end{cases} \\
\ 
Q = &\left\{ (x,y,z) \in \mathbb{R}^3 \, | \: z\ge 0\right\}&
\end{aligned}
\right\}
\implies
\left\{
\begin{aligned}
&\begin{cases}
x = \r \sin \phi \cos \theta, \\
y = \r \sin \phi \sin \theta, \\
z = \r \cos \phi, 
\end{cases}
\\ 
&\ \ \: \r \in [0, +\infty), \ \ 
\phi \in \left[0, \frac{\pi}{2}\right], \ \ 
\theta \in [0, 2\pi)
\end{aligned}
\right.
$$
Note that polar coordinate system 
$$ 
\begin{aligned}
x^2 + y^2 + z^2 & = 
\r^2 \sin^2\phi\cos^2 \theta + \r^2 \sin^2\phi\sin^2 \theta + \r^2\cos^2\phi = 
\\ & = \r^2 \left( \left(\cos^2 \theta + \sin^2 \theta \right) \sin^2 \phi +  \cos^2 \phi\right) 
\\ & = \r^2 \left( \sin^2 \phi +  \cos^2 \phi\right)  = \r^2.
\end{aligned}
$$
Therefore 
$$
\iiint_Q \frac{1}{x^2 + y^2 + z^2} = 
\int_{z=0}^{+\infty}  \int_{y = -\infty}^{+\infty} \int_{x=-\infty}^{+\infty}  \frac{dx\, dy\, dz }{x^2 + y^2 + z^2} 
= 
\int_{\theta=0}^{2\pi}  \int_{\phi=0}^{\frac{\pi}{2}}  \int_{\r=0}^{+\infty} 
\frac{\r^2 \sin \phi}{\r} \, d\r \, d\phi \, d\theta = 
\\
= 
\int_{\theta=0}^{2\pi}  \int_{\phi=0}^{\frac{\pi}{2}}  \int_{\r=0}^{+\infty} 
\!\!\big(\r \sin \phi \big)\, d\r \, d\phi \, d\theta,
$$
where $\r^2 \sin \phi$ is Jacobian.
Hope you can pick it up from here.
