Triple integral of $e^{{(x^2+y^2+z^2)}^\frac32}$

Question

I'm trying to find the result of

$$ \iiint_{R} e^{(x^2+y^2+z^2)^{3/2}} \,dz\,dy\,dx $$

with

$$ R = \left\lbrace (x,y,z) \in \mathbb{R}^3 \mid -1 \le x \le 1, 0\le y \le \sqrt{1-x^2}, 0\le z \le \sqrt{1-x^2-y^2} \right\rbrace. $$ I'm pretty sure I need to use polar coordinates, but I'm stuck with $e^{(x^2+y^2+z^2)^{3/2}}$ , I hope I made my self clear, sometimes it's hard for me to explain in english what i'm trying to do.

Answer 1

Transformation to spherical coordinates is indeed the way to proceed. We have

$$\begin{align} \int_R e^{(x^2+y^2+z^2)^{3/2}}\,dx\,dy\,dz&=\int_0^\pi \int_0^{\pi/2}\int_0^1 e^{r^3}\,r^2\,dr\,\sin(\theta)\,d\theta\,d\phi\\\\ &=\pi \int_0^{\pi/2}\sin(\theta)\,d\theta\,\int_0^1 r^2e^{r^3}\,dr\\\\ &=\frac{\pi (e-1)}{3} \end{align}$$

Answer 2

In polar coordinates, the integration domain is a quarter of a unit ball (an orange slice where the orange axis is along x). Because the integrand only depends on the radius $r=\sqrt{x^2+y^2+z^2}$, the angular part integrates into $4\pi$ (times a quarter, obviously), so:

$$\pi \int_0^1 e^{-r^3}r^2\,dr$$ This is trivially solved by $u=r^3$.

Answer 3

HINT: Use spherical coordinates $$(x,y,z) = (\rho \cos \theta \sin \varphi, \rho \sin \theta \sin \varphi, \rho \cos \varphi)$$ Remembering the Jacobian your integral become $\int \int \int_R \rho^2 e^{\rho^3} \sin \varphi \textrm{ } d\rho d\theta d\varphi$

Answer 4

You should use the parametrization

$$ f(\rho, \phi,\theta) = \begin{cases} x = \rho\cos\phi\sin\theta \\ y = \rho\sin\phi\sin\theta \\ z = \rho\cos\theta \end{cases} $$

with $\rho \in [0,1]$, $\theta \in [0,\pi]$ and $\phi \in [0,\tfrac{\pi}{2}]$ and Jacobian

$$ \rho^2\sin\theta. $$

Thus, your integral becomes

$$ \int_0^1 \int_0^\pi \int_0^\frac{\pi}{2} e^{\rho^3}\rho^2\sin\theta\: \mathrm{d}\theta \mathrm{d}\phi \mathrm{d}\rho = \pi \left(\frac13e^{\rho^3}\right)_0^1\left(-\cos\theta\right)_0^\frac{\pi}{2} = \pi\frac{e-1}{3}. $$