# Show that a triple integral = pi/4

1. Mar 14, 2012

### hm8

1. The problem statement, all variables and given/known data

Show that

$\int\int\int \sqrt{x^{2}+y^{2}+z^{2}}$ $e^{-({x^{2}+y^{2}+z^{2}})} dxdydz = \pi/4$ where the bounds of x, y, and z are 0 to infinity

(The improper integral is defined as the limit of a triple integral over the piece of a solid sphere which lies in the first octant as the radius of the sphere increases indefinitely).

2. Relevant equations

In spherical coordinates, ρ2 = x2 + y2 + z2
dxdydz = ρ2sin∅ drho dpho dtheta

3. The attempt at a solution

I tried converting to spherical coordinates, which gave me

$\int\int\int ρ^{3}$ $e^{-ρ^{2}} sin\phi d\rho d\phi d\theta$

But I'm not sure what my bounds would be (isn't ρ in relation to theta or phi somehow?) or even if I did, I'm not sure I could integrate it...

2. Mar 14, 2012

### HallsofIvy

Staff Emeritus
Since x, y, z range from 0 to infinity, you are talking about the first octant. To cover all four quadrants, $\theta$ normally ranges from 0 to $2\pi$. To cover just the first quadrant, it ranges from 0 to $\pi/2$. To cover all values of z, $\phi$ normally ranges from 0 to $\pi$, to cover just z> 0 it must vary from 0 to $\pi/2$.

3. Mar 15, 2012