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Show that a triple integral = pi/4

  1. Mar 14, 2012 #1


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

    Show that

    [itex]\int\int\int \sqrt{x^{2}+y^{2}+z^{2}} [/itex] [itex] e^{-({x^{2}+y^{2}+z^{2}})} dxdydz = \pi/4[/itex] 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

    [itex]\int\int\int ρ^{3} [/itex] [itex] e^{-ρ^{2}} sin\phi d\rho d\phi d\theta[/itex]

    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. jcsd
  3. Mar 14, 2012 #2


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    Science Advisor

    Since x, y, z range from 0 to infinity, you are talking about the first octant. To cover all four quadrants, [itex]\theta[/itex] normally ranges from 0 to [itex]2\pi[/itex]. To cover just the first quadrant, it ranges from 0 to [itex]\pi/2[/itex]. To cover all values of z, [itex]\phi[/itex] normally ranges from 0 to [itex]\pi[/itex], to cover just z> 0 it must vary from 0 to [itex]\pi/2[/itex].
  4. Mar 15, 2012 #3


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    What about p?
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