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Math Help - Evaluating the iterated integral by converting to polar coordinates.

  1. #1
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    Evaluating the iterated integral by converting to polar coordinates.

    http://www.webassign.net/latexImages...38b1a15e78.gif

    Please click to see the problem. The answer is 16 but I keep getting 3pi

    f(x,y) = 9sqrt(x^2+y^2)
    f(rcos, rsin) = 9sqrt(r^2) = 9r

    From the limits we have 0 =< y =< sqrt(2x-x^2)

    and 0 =< x =< 2

    sqrt(2x-x^2) is a semi-circle in the first quadrant with center (1,0) and r = 1

    so my iterated polar integral is:

    int[(0,pi)int(0,1) 9r^2 drdtheta]

    integrating first with respect to r gives me

    3r^3](0,1) = 3

    then int(0,pi) 3 dtheta = 3pi ... but the answer is 16 so i have no idea what i'm doing incorrectly.

    Thanks in advance!
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  2. #2
    Senior Member
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    Atlanta, GA
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    Your semi-circle cannot be expressed as 0<\theta<\pi, 0<r<1 -- that would be a semi-circle centered on the origin. Starting with y<\sqrt{2x-x^2}, this can be expressed in polar coordinates as r<2\cos\theta. This is your correct upper bound for r.
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