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Help with a triple integral in spherical coordinates

  1. Nov 3, 2012 #1
    The problem statement, all variables and given/known data
    Use spherical coordinates.

    Evaluate[itex]\int\int\int_{E}(x^{2}+y^{2}) dV[/itex] where E lies between the spheres x2 + y2 + z2 = 9 and x2 + y2 + z2 = 25.



    The attempt at a solution

    I think my problem may be with my boundaries. From the given equations, I work them out to be:

    ρ = 3 to 5
    φ = 0 to π
    θ = 0 to 2π

    This gives me the triple integral [itex]\int^{2\pi}_{0}\int^{\pi}_{0}\int^{5}_{3}(x^{2}+y^{2}) dV[/itex] which becomes [itex]\int^{2\pi}_{0}\int^{\pi}_{0}\int^{5}_{3}[/itex](ρ sinφ cos θ)2+(ρ sinφ sin θ)2 ∂ρ∂φ∂θ


    Is this integral correct?
     
  2. jcsd
  3. Nov 3, 2012 #2

    tiny-tim

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    Hi Kaldanis! :smile:

    your limits look ok, but dV is not dρdφdθ, is it? :wink:

    (and why are you using "∂" ? :confused:)
     
  4. Nov 3, 2012 #3
    Hi tiny-tim :smile:

    Should it be dρdθdφ? I've been trying to work through and learn from my book but it's very difficult to understand. I'm having trouble deciding which order to integrate things. Also I've noticed people using r instead of ρ but I don't think that matters to much, it's still from 3 to 5.

    (I had no reason for using ∂, it was just close to ρ! I won't use it again on here:shy:)
     
  5. Nov 3, 2012 #4

    tiny-tim

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    Hi Kaldanis! :smile:
    No.

    You need to learn:

    the volume element in spherical coordinates is dV = r2sinθ drdθdφ …​

    see http://en.wikipedia.org/wiki/Spheri..._and_differentiation_in_spherical_coordinates :smile:

    (and of course you also need to learn the volume element in cylindrical coordinates)
    the order of integration never matters (though usually one order is easier than the others!)

    and yes, i use r for both spherical and cylindrical coordinates, but i believe a lot of people use r for spherical and ρ for cylindrical

    ρ for spherical is fairly uncommon, but as you say it doesn't really matter (but r is easier to type! :biggrin:)
     
  6. Nov 3, 2012 #5
    Multiplication is commutative, so your differential quantities can be multiplied and any order and will still be correct. You're missing some non-differential terms out front of your expression.

    Edit: Ninja'd
     
  7. Nov 3, 2012 #6
    Thank you both for explaining in detail, especially Tim. I'll go study it some more and come back with an answer for you to check. :)
     
  8. Nov 3, 2012 #7
    I don't think this is right. It's a lot to write out in latex so I'll summarise it!

    [itex]\int^{\pi}_{0}\int^{2\pi}_{0}\int^{5}_{3}[/itex] [(r sinφ cosθ)2+(r sinφ sinθ)2]r2sinθ drdθdφ

    = [itex]\int^{\pi}_{0}\int^{2\pi}_{0}[/itex] (2882/5)sin(θ)sin2(φ) dθdφ

    This equals 0 so the final answer is 0? I always doubt it when I calculate a volume to be 0.
     
  9. Nov 3, 2012 #8
    Everything seems ok to me.

    Why do you say that equals 0?
     
    Last edited: Nov 3, 2012
  10. Nov 3, 2012 #9
    If the first part is definitely correct, then I have to integrate [itex]\int^{\pi}_{0}\int^{2\pi}_{0}[/itex] (2882/5)sin(θ)sin2(φ) dθdφ. Evaluating (2882/5)sin(θ)sin2 from 0 to 2π gives me zero, which means the final integral is also zero. Since the answer isn't zero I think my beginning equation must be incorrect? :frown:
     
  11. Nov 3, 2012 #10
    [Edited out various confusions]

    ...

    I think I found the issue. Your definition of dV is inconsistent with your angle definitions. It should be [itex]dV=r^2 sin\phi\ dr\ d\theta\ d\phi[/itex]
     
    Last edited: Nov 3, 2012
  12. Nov 3, 2012 #11
    Thank you! I was just talking with a friend and he also pointed out that I was using theta when I should be using phi. Now I have the correct answer and understand where I messed up.

    :smile:
     
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