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Homework Help: Surface Integrals: What am I doing wrong?

  1. Jan 26, 2010 #1
    Problem: evaluate the double integral of yz over that part of the plane z=y+3 which is inside the cylinder x2+y2=1

    I evaluated with respect to z, from z=0 to z=y+3

    I got (y3+9y+6y2)/2. Then I integrated this over x2+y2=1. To do that, I switched to polar coordinates, letting x=rcos(theta), y=2sin(theta)

    To set up the integral, I used the formula: double integral of [f(x,y,z) dS] = double integral of [f(R(u,v))(multitude of cross product of derivative of R with respect to u, and derivative of R with respect to v) dA] where R is a vector

    In my case, R = rcos(theta)i + rsin(theta)j
    and u=r, and v=theta

    I got (multitude of cross product of derivative of R with respect to u, and derivative of R with respect to v) = r

    Then I integrated [r3sin3(theta) +9sin(theta)+3r2sin2(theta)]r r dr d(theta) over r from 0 to 1, and theta from 0 to 2pi

    My final answer was 3pi/5. The correct answer is 21/2pi/4

    What did I do wrong?
     
    Last edited by a moderator: Jan 26, 2010
  2. jcsd
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