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allsmiles

  • 2 years ago

Evaluate ∫∫ over D (cos(sqrt(x^2+y^2)) dA by changing polar coordinates where the disk is with center of origin and radius 2

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  1. allsmiles
    • 2 years ago
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    I got up to ∫∫[0 to 2pi] [0 to 2] cos(sqrt(r)) r dr dt

  2. allsmiles
    • 2 years ago
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    but I can't solve for it! How do integrate r*cos(sqrt(r))

  3. myko
    • 2 years ago
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    \[\int\limits\int\limits_{D}=\int\limits_{0}^{2\pi} d\theta\ \int\limits_{0}^{2}r \cos rdr\]

  4. allsmiles
    • 2 years ago
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    omg you are correct.

  5. allsmiles
    • 2 years ago
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    spent literally 30 mins trying to see what I did wrong. That was quick thanks a lot man

  6. myko
    • 2 years ago
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    yw

  7. myko
    • 2 years ago
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    by the way:int( rcos r) it's done by integration by parts

  8. allsmiles
    • 2 years ago
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    can't I do substitution

  9. myko
    • 2 years ago
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    you can try

  10. allsmiles
    • 2 years ago
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    haha I just did and it didn't work, alright thanks again!

  11. myko
    • 2 years ago
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    if you have problems look here, :) http://en.wikipedia.org/wiki/Integration_by_parts

  12. myko
    • 2 years ago
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    there exacly you example

  13. myko
    • 2 years ago
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    your*

  14. allsmiles
    • 2 years ago
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    Yeah I got it now, thanks a lot for your help!

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