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Polar coordinate integral

  1. Sep 14, 2016 #1
    1. The problem statement, all variables and given/known data
    m244.PNG

    2. Relevant equations
    Average (area) = 1/Area * integrate of polar

    3. The attempt at a solution
    y= r* sin theta
    x= r* cos theta
    r^2 = x^2+y^2

    upload_2016-9-13_21-4-40.png
     
  2. jcsd
  3. Sep 14, 2016 #2

    BvU

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    Why the square root ?
     
  4. Sep 14, 2016 #3
    x^2 + y^2 =r^2... OH, so only r not root r! and other than that, its fine?
     
  5. Sep 14, 2016 #4

    Ssnow

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    Hi, the ##Area(R)=\pi a^{2}## that you must divide, after ##z(x,y)=f(r,\theta)=\sqrt{x^{2}+y^{2}}=r## and not ##\sqrt{r}##... so inside there is ##r^2## ...
     
  6. Sep 14, 2016 #5
    so the solution is (1/Area) double integral r dr d(theta) ?
     
  7. Sep 14, 2016 #6

    Ssnow

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    No, you have another ##r## inside the integral ...
     
  8. Sep 14, 2016 #7
    1/Area* ∫∫ r dr dθ is not correct ...?
     
  9. Sep 14, 2016 #8

    Ssnow

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    the formula say ##\frac{1}{Area(R)}\int\int_{R}f(r,\theta)rdrd\theta## so you have ##f(r,\theta)\cdot r## inside ...
     
  10. Sep 14, 2016 #9
    so f(r, θ) makes a r,
    then 1/Area* ∫∫ r* r dr dθ
     
  11. Sep 14, 2016 #10

    Ssnow

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    now it is ok, :wink:
     
  12. Sep 14, 2016 #11
    thanks buddy!
     
  13. Sep 14, 2016 #12

    BvU

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  14. Sep 14, 2016 #13
    And I got the answer as ⅓ a
     
  15. Sep 14, 2016 #14

    BvU

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    OK it is ...
     
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