Evaluate $ \iiint_S e^{-(x^2+y^2+z^2)^{3/2}} $ for $S=\{(x,y,z):x^2+y^2+z^2≤4, x>0, y>0, z>0\}$. The given region is $S=\{(x,y,z):x^2+y^2+z^2≤4, x>0, y>0, z>0\}$, i.e. the first octant of the sphere with radius 2. The given integral has the form:
$$ \iiint_S e^{-(x^2+y^2+z^2)^{3/2}}. $$
I'm wondering how can I obtain the upper boundaries for the variables for integrals and how to evaluate this in spherical coordinates. 
The answer is $\dfrac{\pi}{6}(1-e^{-8})$.
 A: $S$ is a filled in segment of a sphere of radius $2$. It's important to try to either draw or visualize the set over which you're integrating.
Since $x^2 + y^2 + z^2 = \rho^2$, we have that $0 < \rho \leq 2$. Since $x,y,z>0$ the segment we're interested in lies in the first octant, so $0< \theta < \pi/2$ and $0 < \phi < \pi/2$ as well. This gives you
$$
\iiint_S e^{-(x^2+y^2+z^2)^{3/2}} \, dV =\int_0^{\pi/2} \int_0^{\pi/2} \int_0^2 e^{-(\rho^2)^{3/2}} \rho^2 \sin{\phi} \, d\rho \, d\theta \, d\phi. 
$$
A: By setting $\rho^2=x^2+y^2+z^2$ we have that the purpose is to integrate $e^{-\rho^3}$ over the region given by the intersection of a solid sphere centered at the origin with radius $2$ and the positive octant. For any $r\in(0,2)$ the measure of the set $\{x^2+y^2+z^2=r^2,(x,y,z)>0\}$ is $\frac{\pi}{2}r^2$, hence by integrating along shells we get that the original integral equals
$$ \int_{0}^{2}\frac{\pi}{2} r^2 e^{-r^3}\,dr = \left.\frac{\pi}{6} e^{-r^3}\right|_{0}^{2} =\color{red}{\frac{\pi}{6}(1-e^{-8})}$$
and we're done.
