Show that a triple integral = pi/4

[hm8]

Homework Statement

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).

Homework Equations

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

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...

 

[HallsofIvy]

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.

 

[hm8]

What about p?