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Help with a simple cos substitution

  1. Apr 6, 2013 #1
    1. The problem statement, all variables and given/known data

    Hi folks, I am sure this is very simple but there are not enough steps given in this calculation for my simple brain to get from the beginning to the end!

    σ = ∫ (dσ/dΩ) = ∫ r2sin2θ (no integral limits given)

    σ = 2∏r2 ∫ (1 - u2) du (integral from -1 to 1)

    σ = 8∏r2 / 3

    2. Relevant equations

    u = cos θ


    3. The attempt at a solution

    I used Ω = sin θ dθ d∅ and first integrated ∅ from 0 to 2∏ to get

    σ = ∫ dσ = r2 ∫ sin2θ dΩ

    σ = 2∏ r2 ∫ sin2θ sin θ dθ

    Use sin2θ = 1 - cos2θ to get

    σ = 2∏ r2 ∫(1 - cos2θ) sin θ dθ

    Let u = cos θ so du/dθ = - sin θ and dθ = -arcsin θ to get

    σ = 2∏ r2 ∫(1 - u2) sin θ -arcsin θ du

    I think sin and arcsin cancel to give

    σ = 2∏ r2 -∫(1 - u2) du

    σ = 2∏ r2 2u

    From the answer that was given I have the integral limits running from -1 to 1 so the final term becomes [2 - (-2)] = 4 which gives

    σ = 8∏ r2

    Where am I going wrong please?
     
  2. jcsd
  3. Apr 6, 2013 #2
    Oh. I differentiated u instead of integrating. Some days there is not enough coffee in the world . . .
     
  4. Apr 6, 2013 #3

    I like Serena

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    Homework Helper

    Hi ck99! :smile:

    You seem to mix up differentials a bit here and elsewhere.

    I'm going to assume you actually intended:

    σ = ∫ dσ = ∫ r2sin2θ dΩ


    Here's another mix-up.
    Differentials always need to balance.

    This is not the case in dθ = -arcsin θ, but then, you don't need it.

    Let's stick to: u = cos θ so du = - sin θ dθ.


    That should be:

    σ = 2∏ r2 ∫(1 - u2)(-du)



     
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