Prove the improper triple integral equals 2*Pi

In summary, the homework statement is trying to solve for the coordinates of a point on a sphere. This can be done by converting to spherical coordinates and integrating by parts. However, when trying to find the limit at infinity, the equation becomes indeterminate.
  • #1
alanthreonus
10
0

Homework Statement



Show that

[tex]\int^{infinity}_{-infinity}\int^{infinity}_{-infinity}\int^{infinity}_{-infinity}sqrt(x^2+y^2+z^2)e^-^(^x^2^+^y^2^+^z^2^)dxdydz = 2\pi[/tex]

Homework Equations



[tex]x^2+y^2+z^2 = \rho^2 [/tex]

The Attempt at a Solution



I converted to spherical coordinates to get

[tex]\int^{2\pi}_{0}\int^{2\pi}_{0}\int^{infinity}_{0}\rho^3e^-^\rho^2sin\phi d\rho d\phi d\theta[/tex]

But I don't think I can integrate that. Am I approaching this the right way?
 
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  • #2
The bounds are independent.
You can get rid of the (phi) and (theta) integrals (ie factor them out).

As for the remaining integrand, you have a power of (rho) multiplied by something you can integrate (may take a second thought to get this 'something' right).
What method does this suggest?
Consider:
(int) x² . x exp(-x²) dx
 
  • #3
OK, integration by parts gives

(-1/2e^(-rho^2))(rho^2+1)

Which is evaluated from 0 to infinity, but unless I'm missing something, the limit at infinity is undefined.

And something else occurred to me. If I integrate with respect to phi, I'll get -cos(phi), which evaluated from 0 to 2*pi is 0. So does that mean I messed up the conversion to spherical coordinates?
 
  • #4
Kinda… In spherical coordinates, what you call phi goes from 0 to Pi, not 2*pi :)
As for the integration over rho, it definitely converges as rho^3 Exp[-rho^2] is 0 at 0 and goes to 0 at infinity (the exponential is stronger than any power of rho) :)
 
  • #5
alanthreonus said:
OK, integration by parts gives

(-1/2e^(-rho^2))(rho^2+1)

Which is evaluated from 0 to infinity, but unless I'm missing something, the limit at infinity is undefined.
You can put the exponential in the denominator and use L'Hopital's rule to evaluate the limit at infinity.
 
  • #6
Thanks a bunch guys, I got it now. And now I feel dumb for not seeing all these things I missed.
 
  • #7
I'm kind of lost. I did all the work up to the very end and have just the limit function left but can't get it to work out.

Can someone check this and let me know what else I need to do or what I messed up on?

I have:

2[tex]\pi[/tex] * lim (t [tex]\rightarrow[/tex] [tex]\infty[/tex] [ -e-t^2/2 * (t2 + 1) + 1/2 ]

I believe that the way I have this written down it is equal to some undefined number because we find that the exponential part goes to 0 while t2 becomes [tex]\infty[/tex]

Thanks in advance, Travis
 
  • #8
[itex]0\cdot\infty[/itex] is an indeterminate form. You can use L'Hopital's rule to evaluate the limit.
 
  • #9
Now I really feel dumb... like the person who started the thread I messed up my limits. For some reason I thought phi was [0, pi/2] so when I finished everything I came up with just pi as a result. Once I fixed this it worked out to 2*pi
 

1. What is an improper triple integral?

An improper triple integral is an integral that involves three variables and has one or more of the limits of integration being infinite or the function being integrated being undefined at certain points.

2. How do you prove that an improper triple integral equals 2*Pi?

To prove that an improper triple integral equals 2*Pi, you would need to evaluate the integral by breaking it into smaller integrals with finite limits and then using the appropriate techniques to solve each integral. Once you have evaluated all the smaller integrals, you can add them together to get the final result of 2*Pi.

3. What is the significance of the value 2*Pi in an improper triple integral?

In an improper triple integral, the value 2*Pi represents the total volume of the region being integrated over. It is a constant value that is commonly used in calculations involving circles, spheres, and other geometric shapes.

4. Can an improper triple integral have a value other than 2*Pi?

Yes, an improper triple integral can have a value other than 2*Pi. The value of an improper triple integral depends on the function being integrated, the limits of integration, and the techniques used to evaluate the integral. In some cases, the value may be a multiple of 2*Pi or a completely different value.

5. Are there any practical applications of proving an improper triple integral equals 2*Pi?

Yes, there are many practical applications of proving an improper triple integral equals 2*Pi. For example, it can be used in physics to calculate the moment of inertia of a solid object, in engineering to determine the volume of a complex shape, and in mathematics to solve differential equations involving spherical coordinates.

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