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How would you do this?

  1. Oct 25, 2013 #1
    How would you go about solving this?
     

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  2. jcsd
  3. Oct 25, 2013 #2

    arildno

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    Dearly Missed

    Integrate it?
    Where is your specific problem?
     
  4. Oct 25, 2013 #3

    UltrafastPED

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    It's a triple integral (a volume, for example). You need two more sets of integration limits. Since one of the integration variables doesn't appear in the integrand you can integrate it directly - it would most likely end up as a factor of 2pi. Looking at the remaining parts, you can integrate each separately, taking into account the integration limits.
     
  5. Oct 26, 2013 #4
    c∫d α∫β 1∫2 ρ^2 sin∅ dρd∅dθ

    = c∫d α∫β [(ρ^3)/3 sin∅] [1,2] d∅dθ
    = c∫d α∫β 7/3 sin∅ d∅dθ
    = c∫d [-7/3 cos∅] [α,β] dθ
    = c∫d [-7/3 cosβ] - [-7/3 cosα] dθ
    = c∫d 7/3 (cosα - cosβ) dθ
    = [(7θ)/3 (cosα - cosβ)] [c,d]
    = [(7d)/3 (cosα - cosβ)] - [(7c)/3 (cosα - cosβ)]
    = (1/3)(7d-7c)(cosα - cosβ)

    This is the general solution where α≤∅≤β and c≤∅≤d
     
  6. Oct 26, 2013 #5
    Sorry, cropped too much there.
     

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  7. Oct 26, 2013 #6

    HallsofIvy

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    That's easy then! Since the limits of integraton are all constants and the integrand is just a product of functions of the separate variables, that is just
    [tex]\left(\int_1^2 \rho^2 d\rho\right)\left(\int_{\pi/2}^\pi \sin(\phi) d\phi\right)\left(\int_0^2 d\theta\right)[/tex]

    Though I must say the limits of integration on that last, [itex]d\phi[/itex], integral look suspicious to me! It's not impossble- but "0 to 2 radians" seems peculiar. "0 to [itex]2\pi[/itex]" would see much more reasonable. That would be integrating over a hemi-spherical "cup", below the xy-plane with thickness from 1 to 2.
     
  8. Oct 26, 2013 #7
    Why is that weird?
     
  9. Oct 26, 2013 #8
    Dude, you should seriously learn latex! It's not that hard and it makes it so much easier for everyone to read what you write. It also looks way better and it's probably better for you as well!

    Just take a look at this link.
     
  10. Oct 26, 2013 #9
    You're integrating using spherical coordinates. ##\theta## is an angle and it seems weird that it goes from ##0## to ##2## radians.
     
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