Evaluate the integral using spherical coordinates Given the integral  $\int^{1}_{0}\int^{\sqrt{1-x^{2}}}_{0}\int^{\sqrt{1-x^{2}-y^{2}}}_{0} \dfrac{1}{x^{2}+y^{2}+z^{2}}dzdxdy$
I need to evaluate this using spherical coordinates.
So far I have that $0\leq r \leq 1$ and I understand that $\theta$ is the angle made in the xy plane and has to be less than or equal to $2\pi$ and $\varphi$ is the angle made revolving around the z-axis and is less than or equal to $\pi$ however I am not sure on how to workout the limits of $\theta$ and $\varphi$ for this question.
 A: In spherical coordinates the function you are integrating over is $$\frac{1}{x^2+y^2+z^2}=\frac{1}{r^2},$$ and  this function is constant over shells of radius $r$. Since the region of integration is the ball of radius $1$ in the first octant, and since the surface area of a shell of radius $r$ is $4\pi r^2$, your integral equals $$\frac{1}{8}\int_0^1\frac{1}{r^2}\cdot 4\pi r^2dr=\frac{\pi}{2}.$$
A: I will work on this problem when radius $r$ is constant in general.
Step 1. Spherical coordinates means:
$x = r \sin(\varphi) \cdot \cos(\theta)$
$y = r \sin(\varphi) \cdot \sin(\theta)$
$z = r \cos(\varphi)$
(Convince your self by drawing pictures)
Step 2. Change of coordinates needs Jacobian
$|J| = r^2 \cdot \sin(\varphi)$ in this case.
So, $dxdydz = r^2 \cdot \sin(\varphi) drd \varphi d\theta$
Step 3. Calculate
$\int_0^r \int_0^\sqrt{r^2-x^2} \int_0^\sqrt{r^2-x^2-y^2} \frac{1}{x^2 + y^2 + z^2} dzdydx$
= 
$\int_0^{\pi/2} \int_0^{\pi/2} \int_0^r \frac{1}{r^2} r^2 \sin(\varphi) drd \varphi d\theta$
=
$\int_0^{\pi/2} \int_0^{\pi/2} \int_0^r \sin(\varphi) drd \varphi d\theta$
=$r \cdot \pi/2$.
A: Jacobian is 
$$ = \cos \phi\,  r^2 \,dr \, d \phi \, d \theta $$ for volume element conversion to spherical coordinates from rectangular. 
The integrand 
$$=  \dfrac{1}{r^2}$$  which simplifies integral to
$$\int^{\pi/2}_{0} \, \int  ^{\pi/2}_{0} \int^1_{0} dr \, \cos \phi \, d\phi \, d \theta $$
$$ = \pi/2,  $$ 
as the area under half sine wave is unity.
