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Trig substitution step (I think)

  1. Nov 19, 2012 #1
    1. The problem statement, all variables and given/known data
    I'm stuck at an attempt to solve an integration step. I think I'm supposed to trig substitute?

    2. Relevant equations
    http://img685.imageshack.us/img685/9158/unavngivetn.png [Broken]

    It is the second to third equation I'm having a hard time with
    3. The attempt at a solution
    From second equation: (I'm only doing the integral)

    [tex] \int_0^{\pi} \frac{Sin \theta}{[R^2 + Z^2 - 2 R Z Cos \theta]^{1/2}} d \theta[/tex]

    [tex] u = cos \theta[/tex]
    [tex] -du = sin \theta[/tex]

    [tex] \int_{-1}^1 \frac{1}{[R^2 + Z^2 - 2 R Z u]^{1/2}} d u [/tex]

    I flipped the boundaries to remove the minus sign.

    And here I'm stuck. I could use some help :b
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Nov 19, 2012 #2

    Dick

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    Try ## u=R^2 + Z^2 - 2 R Z Cos \theta ## Everything except for theta is a constant.
     
    Last edited by a moderator: May 6, 2017
  4. Nov 19, 2012 #3

    Mark44

    Staff: Mentor

    An ordinary substitution will work. Let u = R2 + Z2 - 2RZ cos(θ).
     
    Last edited by a moderator: May 6, 2017
  5. Nov 19, 2012 #4
    First, thanks. I've tried using the whole expression in a substitution, but I got stuck again:

    ## u=R^2 + Z^2 - 2 R Z Cos \theta ##

    ## \frac{du}{d \theta}=2 R Z Sin \theta ##

    ## \frac{du}{2 R Z}= Sin \theta d \theta##

    [tex]\int_{R^2 + Z^2 - 2 R Z}^{R^2 + Z^2 + 2 R Z} \frac{1}{u^{1/2}} \frac{1}{2 R Z} d u [/tex]

    [tex]\frac{1}{2 R Z} \int_{R^2 + Z^2 - 2 R Z}^{R^2 + Z^2 + 2 R Z} \frac{1}{u^{1/2}} d u [/tex]

    [tex]\frac{1}{2 R Z} [2 (R^2 + Z^2 + 2 R Z)^{2/3} - 2 (R^2 + Z^2 - 2 R Z)^{2/3}][/tex]

    [tex]\frac{1}{R Z} [(R^2 + Z^2 + 2 R Z)^{2/3} - (R^2 + Z^2 - 2 R Z)^{2/3}][/tex]

    Don't know where to go from here
     
  6. Nov 19, 2012 #5

    Dick

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    Factor the quadratics inside the (2/3) powers. And check that (2/3). Where did that come from?
     
  7. Nov 19, 2012 #6
    Oh they were supposed to be 3/2 lol (From the integration of (random)^1/2). Anyways will try that
     
  8. Nov 19, 2012 #7

    Dick

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    3/2 doesn't sound right either. You are integrating u^(-1/2).
     
  9. Nov 19, 2012 #8
    Your last comment did the job. Thank you ^^
     
  10. Nov 19, 2012 #9
    Oh there seems to be one more thing I don't quite get. This whole problem was how to calculate the potential of a hollow sphere

    With the equation like this:
    [tex]\phi = \frac{Q}{8 \pi \epsilon_0} \frac{|R + Z| - |R - Z|}{R Z}[/tex]

    It seems that you are able to remove the absolute values when specified when in the sphere or when outside.

    For example when inside Z < R and the equation becomes:

    [tex]\phi = \frac{Q}{8 \pi \epsilon_0} \frac{R + Z - (R - Z)}{R Z}[/tex]

    and when Z > R the equation becomes:

    [tex]\phi = \frac{Q}{8 \pi \epsilon_0} \frac{R + Z + (R - Z)}{R Z}[/tex]

    Do you know why that is? (Why we can just remove the absolute value?)
     
  11. Nov 19, 2012 #10

    Dick

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    It's just because |x|=x if x>=0 and |x|=(-x) if x<0. If you know the sign of x you can replace |x| with a simpler expression.
     
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