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Surface integral from vector calculus

  1. Apr 30, 2009 #1
    1. The problem statement, all variables and given/known data
    F = (x^2) i + (y^2) j + (z) k, S is the cone z = (x^2+y^2) ^ (1/2), with x^2 +y^2 <= 1, x >= 0, y >= 0, oriented upward.


    2. Relevant equations
    All of the above


    3. The attempt at a solution

    My attempted solution is 0. But other students claim that the answers is (2/3)pi.

    My work is this:
    double integral ( (x^2) i + (y^2) j + (z) k ) . ( -(x / (x^2+y^2)^ (1/2) i -(y / (x^2+y^2)^ (1/2) j + (x^2 + y^2) ^ (1/2) k ) dx dy

    Then finding common denominator and simplifying:

    double integral ( (- x - y ) / (x^2+y^2)^(1/2) ) dx dy

    Changing to polar coordinates:

    (first integral 0 to 2pi)(second integral 0 to 1) (-cos(theta) - sin(theta) / (r^2) ) * rdrd(theta)

    Finally, doing those integrals I have the result of 0.
     
  2. jcsd
  3. Apr 30, 2009 #2

    HallsofIvy

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    I do not get 0 as the integral.
     
  4. Apr 30, 2009 #3

    tiny-tim

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    Welcome to PF!

    Hi m453438! Welcome to PF! :smile:

    (have a theta: θ and a pi: π and a √ and a ∫ and a ≤ and ≥, and try using the X2 tag just above the Reply box :wink:)

    Useful tip: when you do it again, try putting r = √(x2 + y2) at the start

    it makes it a lot easier to write, which means you're less likely to make a mistake. :wink:
     
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