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Homework Help: Stokes' Theorem

  1. Mar 4, 2007 #1
    Here's my problem:
    Take u=(x^3)+(y^3)+(z^2) and v=x+y+z and evaluate the surface integral
    double integral of grad(u) x grad(v) ndS where x is the cross product and between the cross product and the ndS there should be a dot product sign. The region S is the hemisphere x^2+y^2+z^2=1 with z greater than or equal to 0 and n has non-negative k component.

    Here's my work:
    grad(u)=(3x^2)i-(3y^2)j+(2z)k
    grad(v)= i+j+k
    grad(u) x grad (v)= (-3y^2-2z)i+(2z-3x^2)j+(3x^2-3y^2)k

    curl F * k = (-6x+6y)
    Thus I=double integral of curl F * ndS = double integral of curl F *k dA = double integral of (-6x+6y)dA.

    I used polar coordinates:
    -6 integral from 0 to 2pi integral from 0 to 1 (rcost-rsint) rdrdt. However, this integral equals zero. Is this right?

    Is this flux zero?
     
  2. jcsd
  3. Mar 4, 2007 #2

    cristo

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    Is this what you mean: [tex]\int\int\nabla u \times\nabla v \cdot \bold{n}dS [/tex]?

    Where does the minus sign in front of the j component come from? [tex]\nabla u=\frac{\partial u}{\partial x}\bold{i}+\frac{\partial u}{\partial y}\bold{j}+\frac{\partial u}{\partial z}\bold{k}[/tex]
    These look correct with the amended form for grad(u)

    What's F? You should define anything you introduce. I imagine you've used Stokes' Theorem here, but where?

    [As an aside, it would be helpful if you could learn LaTex; it's very simple to learn! Click on one of the images in my text, and follow the link to the tutorial]
     
    Last edited: Mar 4, 2007
  4. Mar 4, 2007 #3

    Dick

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    I think I would agree the total flux is zero, but I don't see how you are getting it. What does the stated problem have to do with Stokes Theorem? Where did the curl come from? What is F?
     
  5. Mar 4, 2007 #4
    Yes, the integral that cristo wrote is what I'm looking for.

    Can't I just let F=grad(u) x grad (v)= (-3y^2-2z)i+(2z-3x^2)j+(3x^2-3y^2)k?

    I made a mistake. u should equal (x^3)-(y^3)+(z^2). That's why I have a negative j component.

    If I let F equal the above, isn't this correct?
     
  6. Mar 4, 2007 #5

    Dick

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    You are integrating F. Not curl(F). So you can't take it to a line integral.
     
  7. Mar 4, 2007 #6

    Dick

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    See my post to your latest question. grad(u)xgrad(v) is the curl of something. What?
     
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