Volume of a three dimensional gaussian

[LizardCobra]

Homework Statement

How can I find the volume of a three dimensional gaussian exp\left [ \frac{-x^2}{\sigma_{x}} \frac{-y^2}{\sigma_{x}}\frac{-z^2}{\sigma_{z}} \right ] ? Since it is a gaussian, the volume should actually extend to infinity. It seems like there should be a simple double or triple integral, but I can't figure out how to set it up.

Homework Equations

The Attempt at a Solution

I multiplied the integral of each gaussian over all space (and of course each of these integrals converges)
V = \int_{-\infty}^{\infty}exp\left [ \frac{-x^2}{\sigma_{x}} \right ]\int_{-\infty}^{\infty}exp\left [ \frac{-y^2}{\sigma_{y}} \right ]\int_{-\infty}^{\infty}exp\left [ \frac{-z^2}{\sigma_{z}} \right ], but I'm not sure if this is right.Thanks

 

[mfb]

That should be correct.
The volume (as seen in a 4-dimensional space) extends to +- infinity in all directions, but its height (=function value) drops so quickly that the total volume remains finite.

 

[Dick]

Hopefully you have a sum in the exp and not a product.

 

[LizardCobra]

Yes, that was meant to be a sum, not a product.

How is the volume in 4D space? I had wanted to find the volume in 3D space, which should only requires a double integral. But I was not sure how to account for the density of the function changing in all three directions.

 

[Ray Vickson]

Yes, that was meant to be a sum, not a product.

How is the volume in 4D space? I had wanted to find the volume in 3D space, which should only requires a double integral. But I was not sure how to account for the density of the function changing in all three directions.

No: volume in 3d requires a 3-dimensional integral (see any calculus textbook). Anyway, I think the use of the word "volume" is unnecessary and maybe misleading: you just need to compute an integral of some function over some 3-dimensional region. Period.

 

[Dick]

Yes, that was meant to be a sum, not a product.

How is the volume in 4D space? I had wanted to find the volume in 3D space, which should only requires a double integral. But I was not sure how to account for the density of the function changing in all three directions.

You don't find the volume of a function f(x,y,z), you think of it as a mass density and find the total mass by integrating dxdydz.

 

[LizardCobra]

That makes much more sense. Thank you!