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Trouble with surface integral

  1. Mar 27, 2013 #1

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    1. The problem statement, all variables and given/known data
    Given is the vector field, [itex]\overline{A}[/itex] = (x2-y2, (x+y)2, (x-y)2). The surface: [itex]\overline{B}[/itex] = (u+v, u-v, uv). The restrictions are the following: -1≤ u, v≤ 1, and the z-component of the normal has to be positive.
    Calculate I, I = ∫∫[itex]\overline{A}[/itex][itex]\cdot[/itex][itex]\overline{n}[/itex]dS

    2. Relevant equations




    3. The attempt at a solution
    What I did was I first tried calculating the normal by using the cross product between the two vectors tangent to the surface. This however got me a normal vector with a z component equal to zero. Then I tried expressing u and v in terms of x and y, plugging that into the parameter for z, z = uv, got me z as a function of x and y. I then defined a potential, Φ = z - f(x,y), where z = f(x,y). Calculating n=∇Φ got me a normal vector with non-zero z-component. Easy.

    Next I calculated the dot product between the vector field, A, and the normal, n, in terms of u and v. This is where I got stuck. I'm not sure how to calculate the integral given the boundaries above.

    -1≤ u, v≤ 1 is the same as -1≤ u< ∞, -∞< v ≤ 1. I tried integrating with respect to one of the variables but what happens with the infinites when evaluating the result?

    Maybe I missed something, my calculus skills are a little bit rusty at the moment.

    Oh, I was wondering, how come the normal vectors calculated from the cross product and gradient are different? aren't they supposed to be the same?
     
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  3. Mar 27, 2013 #2

    LCKurtz

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    Show us your steps. I don't get zero in the 3rd component.

    You want to leave it all in terms of u and v. And use the formula$$
    \iint_S \vec F\cdot \hat n dS = \pm\iint_{(u,v)}\vec F(u,v)\cdot \vec R_u\times \vec R_v\, dudv$$using the appropriate sign to agree with your orientation (plus if ##\vec R_u\times \vec R_v## is in the right direction).
     
  4. Mar 29, 2013 #3

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    I calculated the normal again and got it to be:
    n = (-(u + v), u - v, 2).
    This has a z-component > 0.

    Then I expressed A in terms of u and v:
    A = (uv, u2, v2).

    The dot product of A and n is:
    4(2v2 + u3 - 2u2v - uv2)

    I can't see how the integral of this is going to be finite, given the boundaries.
     
  5. Mar 29, 2013 #4

    vela

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    I think the problem intends for you to take ##-1 \le u \le 1## and ##-1 \le v \le 1##.
     
  6. Mar 29, 2013 #5

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    Ahhh, you're right!

    I thought the comma was there to separate.

    Thanks!
     
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