A triple integral over the unit ball I'm trying to find the following triple integral over the unit ball.
$$\iiint_{x^2 + y^2 + z^2 \leq 1} e^{(1-x^{2}-y^{2})^{3/2}} {\rm d} x \, {\rm d} y \, {\rm d} z$$
I am able to find a suitable parametrization but I can't get the bits in the exponent in terms of $r$.  Any hints would be appreciated.
 A: Hint. Use cylindrical coordinates, integrating with respect to $z$ on the inside.
A: The integral boils down to
$$\begin{aligned} I &= \iiint ^{\sqrt{1-\left( x^{2} +y^{2}\right)}}_{-\sqrt{1-\left( x^{2} +y^{2}\right)}} e^{\left( 1-x^{2} -y^{2}\right)^{\frac{3}{2}}} {\rm d} z \, {\rm d} x \, {\rm d} y\\ &= 2\iint e^{\left( 1-x^{2} -y^{2}\right)^{\frac{3}{2}}}\sqrt{1-\left( x^{2} +y^{2}\right)} {\rm d}x \, {\rm d}y \end{aligned}$$
where the double integral must be calculated over the unit circle. Substitute $x = r \cos \theta$ and $y=r\sin \theta$
$$\begin{aligned} I=\iint e^{\left( 1-r^{2}\right)^{\frac{3}{2}}}\sqrt{1-r^{2}} rdrd\theta \\
=\int ^{1}_{0} e^{\left( 1-r^{2}\right)^{\frac{3}{2}}}\sqrt{1-r^{2}} rdr\cdotp \int d\theta \\ \end{aligned}$$
Consider the integral
$$\begin{aligned}A=\int ^{1}_{0} e^{\left( 1-r^{2}\right)^{\frac{3}{2}}}\sqrt{1-r^{2}} rdr\ \end{aligned}$$
Substitute$$\begin{aligned}\ 1-r^{2} =t^{2}\\
-2rdr=2tdt\\
rdr=-tdt\\ \end{aligned}$$
$$\begin{aligned}{A=\int ^{1}_{0} e^{t}}^{3}\cdotp t^{2} dt=\frac{e-1}{3}\\
So,\ I=4\pi \cdotp \frac{e-1}{3}\\ \end{aligned}$$
Does this help?
