How to calculate an integral I wonder how the integral 
$$\int_{-1}^1 \! \int_0^{\sqrt{1-x^2}} \! \int_0^{\sqrt{1-y^2-x^2}} \! 1 \, dz  \, dy  \, dx $$
Any ideas?
 A: If you really want to do this without geometry or spherical coordinates, here's what it looks like:
$\begin{align}
\int_{-1}^1 \! \int_0^{\sqrt{1-x^2}} \! \int_0^{\sqrt{1-y^2-x^2}} \! 1 \, dz  \, dy  \, dx &= \int_{-1}^1 \! \int_0^{\sqrt{1-x^2}} \! \sqrt{1-y^2-x^2}  \, dy  \, dx \\
&= \int_{-1}^1 \! \int_0^{\sqrt{1-x^2}} \! \sqrt{(1-x^2)-y^2}  \, dy  \, dx
\end{align}$
At this point, you can do the change of variables $y=\sqrt{1-x^2}\sin t$, which gives us $dy=\sqrt{1-x^2}\cos t \, dt$, and changes the limits on the inside integral:
$\begin{align}
\int_{-1}^1 \! \int_0^{\pi/2} \! \sqrt{(1-x^2)(1-\sin^2 t)}\sqrt{1-x^2}\cos t  \, dt  \, dx &= \int_{-1}^1 \! (1-x^2) \int_0^{\pi/2} \! \sqrt{\cos^2 t}\cos t  \, dt  \, dx \\
&= \int_{-1}^1 \! (1-x^2) \int_0^{\pi/2} \! \cos^2 t  \, dt  \, dx \\
&=\int_{-1}^1 \! (1-x^2) \frac{\pi}{4}  \, dx \\
&=\frac{\pi}{4} \int_{-1}^1 \! (1-x^2) \, dx \\
&=\frac{\pi}{2} \int_0^1 \! (1-x^2) \, dx \\
&=\frac{\pi}2\cdot\frac23 = \frac{\pi}3
\end{align}$
Is that pretty much what you're looking for?
