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From Cartesian to spherical integral

  1. Apr 13, 2014 #1
    1. The problem statement, all variables and given/known data
    coords.png


    2. Relevant equations

    [itex][/itex]

    3. The attempt at a solution
    Is this the correct setup?

    [itex]\int^{\pi}_{\frac{3\pi}{4}}\int^{2\pi}_{0}\int^{\sqrt{2}}_{0}\frac{1}{\rho^{2}} rho^2 Sin\phi d\rho d\theta d\phi[/itex]

    I gave up on itex. It was either that or my computer flying out the window. My apologies.
     
    Last edited: Apr 13, 2014
  2. jcsd
  3. Apr 13, 2014 #2

    Simon Bridge

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    You came up with:
    $$\int_{3\pi/4}^\pi \int_0^{2\pi}\int_0^{\sqrt{2}}\frac{\sin\phi\;\text{d}\rho \text{d} \theta \text{d}\phi}
    {\rho^2}$$
    (use the quote button on this post to see how I did that ;) )

    Please show your working.
     
  4. Apr 14, 2014 #3
    God, that looks complicated. I'll dedicate some time to actually learning how to write those integrals as nicely as you do after this week. I'm drowning in chemistry and math right now and haven't got a second to spare to learn itex right now :).

    What I did was I drew a picture and then basically looked at it.
    The limits for theta are obvious. The phi I got from setting x=0 on the upper limit and getting z=-x in the zy plane, this is 45 degree line, it's negative so 3pi / 4.

    As for the rho limits, that's where I get less sure of what I'm doing.

    I think that what I'm looking for is an ice cream cone-shape. Therefore the rho starts from 0 and ... Oh no wait I think I see it now. If I set the z-limits equal to each other I get the intercept which is r=1.
    So then rho would go from 0 to 1. Correct?
     
  5. Apr 14, 2014 #4

    LCKurtz

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    That is correct except why not cancel the ##\rho^2## factors?


    No, ##\rho## isn't the same as ##r##. Your original limits are correct.
     
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