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Homework Help: Question on surface integral of curl

  1. Oct 1, 2007 #1


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

    Let F be [tex]F = ( x^2 z^2 ) i + (sin xyz) j + (e^x z) k.[/tex][tex]Find \int\int \nabla \times F \cdot n dS[/tex]

    where the region E is above the cone [tex]z^2 = x^2 + y^2[/tex] and inside the sphere centered at (0,0,1) and with radius 1. (so it is [tex]x^2 + y^2 + (z-1)^2 = 1).[/tex]. I know that they intersect at z = 1, z = 0. So the boundary is at z = 1.

    2. Relevant equations
    So I know that I can apply Stokes Theorem here or evaluate the surface integral directly.

    If I try Stokes Theorem, I apply it to the boundary [tex]x^2 + y^2 + (z-1)^2 = 1[/tex], which can be parametrized by [tex]r(\theta) = cos(\theta) i + sin(\theta) j + k, r'(\theta) = -sin(\theta) i + cos(\theta) j + 0 k[/tex]

    But when I try [tex]F(r(\theta)) \cdot r'(\theta)[/tex], I get [tex]\int_0^{2 \pi} -sin(\theta) cos(\theta)^2 + cos(\theta) sin( cos(\theta) sin(\theta)). [/tex]. Which is pretty much impossible to evaluate, since I never heard of integrating a sine function within a sine function

    as for trying to evaluate the surface integral directly... I get the cross product which is...

    [tex]\nabla \times [tex]F[/tex][tex] = (-x y cos(xyz))i + (2x^2 z - e^x z)j + (y z cos(xyz) k).[/tex]

    (it should read \nabla \times F but my edit doesn't change the tex...)

    So I try to evaluate the surface integral through the northern hemisphere. Of [tex]x^2 + y^2 + (z-1)^2 = 1)[/tex], I get...

    [tex]\frac{\partial g}{\partial x} = 2x, \frac{\partial g}{\partial y} = 2y[/tex]


    [tex]P \frac{\partial g}{\partial x}- Q \frac{\partial g}{\partial y} + R[/tex]
    [tex]= 2x xy cos(xyz) - 2y 2x^2 z + 2y e^y z + yz cos(xyz)[/tex]... which is like totally impossible to evaluate with a surface integral.

    So what am I doing wrong?
    Last edited: Oct 1, 2007
  2. jcsd
  3. Oct 1, 2007 #2


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    Another question incidental to this:

    div (curl) F = 0. But yet, Stokes Theorem applies to a vector function. Is there any fundamental difference between [tex]\int\int \nabla \times F \cdot dS[/tex] and [tex]\int\int F \cdot dS[/tex] when both are vector functions? Yet, the divergence theorem applies to the second one, when [tex]\int\int F \cdot dS = \int\int\int div F dV[/tex]. If there is no fundamental difference between [tex]\nabla \times F \cdot dS[/tex] and [tex]F \cdot dS[/tex], then why does the divergence theorem work for the latter but not the former? If it works for the former, then [tex]\int\int \nabla \times F \cdot dS = \int\int\int div \nabla \times F = 0[/tex] (but this is clearly not the case, otherwise Stokes' Theorem would be useless. Can anyone please clarify the issue to me? Thanks
  4. Oct 3, 2007 #3


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