Can I solve this integral without spherical coordinates? Def.
$$
f: \; [-r, r]^3 \rightarrow \mathbb{R}, \; (x_1, x_2,x_3) \mapsto \begin{cases}
\sqrt{r^2 - x_1^2 - x_2^2 - x_3^2}, & \text{for } x_1^2 + x_2^2 + x_3^2 \leq r^2, \\
0, & \text{else}.
\end{cases}
$$
with $r > 0$.
I need to prove:
$$
\int_{-r}^r \int_{-r}^r \int_{-r}^r f(x_1, x_2, x_3) \mathrm{d}x_1 \mathrm{d}x_2 \mathrm{d}x_3 = \frac{\pi^2}{4}r^4.
$$
I was able to solve the integral by first transforming $(x_1, x_2, x_3)$ into spherical coordinates $(p, \theta, \varphi)$ but I am still at the point in my studies where I do not know about coordinate transformations of integrals.
My question: Is there a way to calculate the integral without spherical coordinates or is there a simple proof such that I can do it in spherical coordinates?
What I did:
\begin{align}
& \int_{-r}^r \int_{-r}^r \int_{-r}^r f(x_1, x_2, x_3) \mathrm{d}x_1 \mathrm{d}x_2 \mathrm{d}x_3 \\
=& \int_{0}^r \int_{0}^\pi \int_{-\pi}^\pi \sqrt{r^2-p^2} \cdot p^2\sin\theta\; \mathrm{d}\varphi \mathrm{d}\theta \mathrm{d}p \\
=& \dots \\
=& 4\pi \int_0^r \sqrt{r^2-p^2} \cdot p^2 \mathrm{d}p \\
=& \dots \\
=& 4\pi \left[ \frac{1}{8} \arcsin\frac{p}{r} - \frac{1}{8}\sin\left( 4\arcsin\frac{p}{r} \right) \right]_{p=0}^{p=r} \\
=& \frac{\pi^2}{4} r^4
\end{align}
 A: (I'm writing $R$ for your $r$.)
We can use "spherical coordinates in disguise" as follows: Choose a  tagged partition
$$0=r_0<r_1<r_2<\ldots<r_N=R$$
of the interval $[0,R]$ with tags $\rho_i\in[r_{i-1},r_i]$. This setup induces a partition of the ball of radius $R$ into shells of thickness $r_i-r_{i-1}$. We then have the Riemann sum approximations
$$J:=\int_{[-R,R]^3} f(x)\>{\rm d}(x)\approx\sum_{i=1}^N\sqrt{R^2-\rho_i^2}\>4\pi \rho_i^2\>(r_i-r_{i-1})\approx4\pi\int_0^Rr^2\>\sqrt{R^2-r^2}\>dr\ ,$$
whereby the errors implied by $\approx$ can be made arbitrarily small. Therefore the two integrals on the LHS and the RHS are in fact equal. In order to compute the latter we substitute $r:=R\sin\theta$ $0\leq\theta\leq{\pi\over2}$ and obtain 
$$J=4\pi R^4\int_0^{\pi\over2}\sin^2\theta\cos^2\theta\>d\theta=\pi R^4\int_0^{\pi/2}\sin^2(2\theta)\>d\theta={\pi^2\over4} R^4\ .$$
A: There is, but it is sort of an underhanded trick, so you might not like it:
$$\int_{-r}^r \int_{-r}^r f(x_1, x_2, x_3) \;dx_1 \,dx_2$$ is the volume of a half (3-dimensional) sphere with radius $\sqrt{r^2-x^2_3},$ so the full integral is (using a formula known from school mathematics),
$$\frac{2}{3}\pi\int^r_{-r}\left(\sqrt{r^2-x^2_3}\right)^3\,dx_3.$$ Now, it's time to substitute $x_3=r\,\sin\theta$ (you always do that when you see such square root expressions in integrals) to obtain $$\frac{2}{3}\pi\,r^4\int^{\pi/2}_{-\pi/2}\cos^4\theta\,d\theta.$$ After that, you can go through tedious integration by parts, or remember and apply the double angle formula for $\cos$ twice, so that
$$\cos^4\theta=\frac{3}{8}+\frac{1}{2}\cos2\theta+\frac{1}{8}\cos4\theta.$$ Then (the integrals over full periods of $\cos2\theta$ and $\cos4\theta$ are conveniently $=0$), you finally get $$\frac{2}{3}\pi\,r^4\cdot\frac{3}{8}\pi=\frac{\pi}{4}r^4.$$
A: You can at least integrate over one variable using the formula for $\int \sqrt {1-x^2} \, dx $ found here: Integral of $\sqrt{1-x^2}$ using integration by parts
