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Double Integral, where did I go wrong? Related to polar coordinates.

  1. Nov 6, 2011 #1
    ∫∫cos(x^2 + y^2)dA, where R is the region that lies above the x-axis within the circle x^2 + y^2 = 9.

    Answer: .5pi*sin(9)

    My Work:
    ∫(0 ->pi) ∫(0 -> 9) cos(r^2) rdrdθ

    u = r^2
    du = 2rdr
    dr = du/2r

    .5∫(0 ->pi) ∫(0 -> 9) cos(u) dudθ
    .5∫(0 ->pi) sin(u)(0 -> 9) dθ
    .5∫(0 ->pi) sin(r^2)(0 -> 9) dθ
    .5∫(0 ->pi) [sin(9^2) - sin(0)] dθ
    .5∫(0 ->pi) sin(81) dθ
    .5[sin(81) θ](0 -> pi)
    .5pi*sin(81)

    However, the answer is suppose to be .5pi*sin(9). Where did I go wrong? Am I not suppose to square r, and if not then doesn't that mean everything I did below the substitution doesn't work?

    Thank you for your time and patience. I apologize for my integral notation, I'm not sure of a better way to type it.
     
  2. jcsd
  3. Nov 6, 2011 #2

    LCKurtz

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    Your circle has radius 3, not 9. r should go from 0 to 3.
     
  4. Nov 6, 2011 #3
    O.O! I love you forever, thank you!
     
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