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Stokes theorem problem

  1. Sep 29, 2007 #1


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    1. The problem statement, all variables and given/known data
    Let C be the closed curve that goes along straight lines from (0,0,0) to (1,0,1) to (1,1,1) to (0,1,0) and back to (0,0,0). Let F be the vector field [tex]F = (x^2 + y^2 + z^2) i + (xy + xz + yz) j + (x + y + z)k. Find \int F \cdot dr[/tex]

    By Stokes Theorem, I know that I can transform this into a curl (instructions say don't evaluate the line integral directly).

    So I know I get a form of square that goes CCW. Projected on the xz-plane, it is simply a square (but the y-coordinate of the right end is 1).

    I think the plane is described by x = z, since there is no dependence on y. So x-z = 0

    Anyways, I take the curl of F to get

    [tex]F \cdot dr = (1-x-y) i + (1-2z) j + (y+z-2y) k.[/tex]

    I take the dot product of this with the plane z - x = 0 to get

    [tex] \nabla f = -i + k [/tex]

    and dot that with F to get
    [tex]F \cdot \nabla f = x + y -1 + y + z - 2y = x + z -1 [/tex]

    so then

    [tex]\int_0^1 \int_0^1 (x + z -1) dx dy.[/tex] Since x = z...

    [tex]\int_0^1 \int_0^1 dx dy = \int_0^1 x^2 - x dy = \frac{1}{3} - \frac{1}{2} = - \frac{1}{6}[/tex]

    am I doing this correctly?
  2. jcsd
  3. Sep 29, 2007 #2


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    Part of your LaTex didn't come through. How do you take the dot product of a vector with a plane?

    And what do you mean by "dot that with F"?

    You have [tex]F \cdot \nabla f [/tex]

    That confuses me greatly. You have a function "F" but say nothing about "f". If you meant [itex]F \cdot \nabla F[/itex], that is not at all what you want.
    Stokes theorem has
    [tex]\int\int \nabla \times \vec{F}\cdot \vec{n}dA[/tex]
  4. Oct 1, 2007 #3


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    Okay, here you have the dot product of a vector with the gradient of the plane:
    (example 2). But this could be confusing, and there is an alternative way to express Stokes Theorem after the dot product has been taking.

    This is Stokes Theorem as a surface integral. (the gradient approach to the surface integral is at http://tutorial.math.lamar.edu/Classes/CalcIII/SurfIntVectorField.aspx )

    As we know,

    [tex]\int\int \nabla \times \vec{F}\cdot \vec{n}dA = \int\int F \cdot dS = \int\int -P \frac{\partial g}{\partial x} - Q \frac{\partial g}{\partial y} + R[/tex]

    [tex]\nabla \times F(x,y,z) = P i + Q j + R k[/tex] and [tex] g = f(x,y) = z[/tex]

    The equation of my plane should be z = f(x,y) = x, right? (the plane is f, and determined by "Let C be the closed curve that goes along straight lines from (0,0,0) to (1,0,1) to (1,1,1) to (0,1,0) and back to (0,0,0).")

    in that case, [tex]\frac{\partial g}{\partial x} = 1, \frac{\partial g}{\partial y} = 0[/tex]

    So correcting my arithmetic..

    [tex]F \cdot \nabla f = -(1 - x - y) + y + z - 2y dA[/tex]
    = [tex]\int_0^1 \int_0^1 -1 + x + z [/tex]

    We take z = x. Is this an appropriate choice? If so

    [tex]\int_0^1 \int_0^1 2x - 1 dx dy = \int_0^1 x^2 - x (1,0) = \int_0^1 0 ....[/tex]

    what am I doing wrong here?
    Last edited: Oct 1, 2007
  5. Aug 6, 2009 #4
    You set [tex]g = f(x,y) = z[/tex], so your D will be the projection of graph g at x-y plane.
    You can substitute [tex]z=x[/tex] at there. I got zero as well for this question.
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