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Evaluate the double integral

  1. Nov 7, 2012 #1
    1. The problem statement, all variables and given/known data

    [itex]\int^{\pi}_{0}[/itex] [itex]\int^{1-sin\theta}_{0}[/itex] r[itex]^{2}[/itex] cos[itex]\theta[/itex] drd[itex]\theta[/itex]


    I keep getting an answer of 0 but i am most certain that i am getting my trig messed up somewhere.
    1/3 [itex]\int^{\pi}_{0}[/itex] r[itex]^{3}[/itex] cos[itex]\theta[/itex][itex]d[itex]\theta[/itex] from 0 to 1-sin\theta

    then i get

    1/3 [itex]\int^{\pi}_{0}[/itex] (1-sin\theta)^3 cos\theta d\theta

    I then use substitution,u for 1-sin\theta then get 1/3 times -1/4(u)^4

    substitute back the 1-sin\theta and evaluate from 0 to \pi and I keep getting zero please help thanks
     
  2. jcsd
  3. Nov 7, 2012 #2

    tiny-tim

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    hi whynot314! :smile:

    looks ok …

    the integrand is anti-symmetric about π/2, so it should be zero :wink:
     
  4. Nov 7, 2012 #3
    thanks, I was just concerned bc these area ones are usually never turn out to be zero thanks.
     
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