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Evaluate a simple double integral

  1. Jul 7, 2009 #1
    Evaluate a "simple" double integral

    1. The problem statement, all variables and given/known data

    Evaluate the double integral of f(x,y) = square root (1 - x^2 - y^2) over the disk centred at the origin of radius 1

    2. Relevant equations



    3. The attempt at a solution
    So the disc of radius one has boundaries x^2 + y^2 = 1
    i am integrating it in the order dydx
    the disk is bound above by y = root (1 - x^2) and below by y = -root (1 - x^2)

    So the inner part of the double integral would be
    Integral from -root (1-x^2) to root (1-x^2) of root (1 - x^2 - y^2) dy

    in a list of integrals i found that if you have an integral in the form root (a^2 - u^2), the antiderivative is u/2 * root (a^2 - u^2) + a^2/2 * sin^-1 (u/a)

    so if i set a^2 = 1-x^2 and u^2 = y^2, my integral has exactly this form, so the solution should be
    root(1-x^2)/ 2 * root (1 - x^2 - (1 - x^2)) + (1-x^2)/2 * sin^-1 (root (1-x^2)/ root (1-x^2) - [ (-root(1-x^2) / x * (root (1 - x^2 - (1 - x^2)) + (1-x^2)/2 * sin ^-1 (-root(1-x^2) / root (1-x^2)

    Which simplifies pretty cleanly to

    (1-x^2) / 2 * (sin^-1 (1) - sin^-1 (-1)

    = (1-x^2)/2 * (-Pi)
    = -Pi/2 + Pix^2/2

    Now for the outer integral, i have to integrate this from x = -1 to x = 1

    and I get
    Integral of -Pi/2 + Pix^2/2 = -Pi x/2 + Pi x^3 /6
    evaluating this from -1 to 1 i get
    - Pi / 2 + Pi/6 - (Pi / 2 - Pi / 6) = -2Pi / 3

    But this gives me a negative value and I thought that the integral should be positive since it is essentally calculating the volume below the curve z = root (1 - x^2 - y^2)...and that function is positive for the entire disk of radius 1 , so i dont see why i am getting a negative number...what in the world is going on :)
     
  2. jcsd
  3. Jul 7, 2009 #2
    Re: Evaluate a "simple" double integral

    Using polar coordinates from the start will simplify the question
     
  4. Jul 7, 2009 #3
    Re: Evaluate a "simple" double integral

    Yeah but we have not yet learned polar coordinates :) so I have no idea how to use them
     
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