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Line Integral in Cylindrical Coordinates

  1. Sep 13, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the value of the (surface) integral [tex]\int curl \textbf{A} \bullet \textbf{a} [/tex]

    if the vector [tex]\textbf{A}=y \textbf{i}+z \textbf{j}+x \textbf{k}[/tex]

    and S is the surface defined by the paraboloid [tex]z=1-x^2-y^2[/tex]


    2. Relevant equations

    [tex]x=s\cos\phi[/tex]

    [tex]y=s\sin\phi[/tex]

    [tex]d\textbf{l}=ds\mathbf{\hat{s}}+s d\phi\mathbf{\hat{\phi}}+dz\mathbf{\hat{z}}[/tex]

    [itex]\mathbf{\hat{x}}=\cos\phi\mathbf{\hat{s}}-\sin\phi\mathbf{\hat{\phi}}[/itex]

    [itex]\mathbf{\hat{y}}=\sin\phi\mathbf{\hat{s}}+\cos\phi\mathbf{\hat{\phi}}[/itex]

    3. The attempt at a solution

    First I used Stokes' theorem in order to turn the integral into a line integral, the integral of the dot product of [tex]\textbf{A}[/tex] and [tex]d\textbf{l}[/tex]

    Then I turned [tex]\textbf{A}[/tex] into cylindrical coordinates using the above x hat and y hat equations. When I took that dot product of [tex]\textbf{A}[/tex] and [tex]d\textbf{l}[/tex], I came up with an answer of:

    [tex](s\cos\phi\sin\phi+z\sin\phi)ds\\\ +\\\ (-s^2\sin^2\phi+z\s\cos\phi)d\phi\\\ +\\\ s\cos\phi dz[/tex]

    I'm trying to reduce it down to one parameter in order to do the integration.

    So, s=1-z. Would that imply ds=-dz? Is dz=1?
     
    Last edited: Sep 13, 2009
  2. jcsd
  3. Sep 13, 2009 #2

    gabbagabbahey

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    You mean [itex]\int_{\mathcal{S}} (\mathbf{\nabla}\times\textbf{A})\cdot d\textbf{a}[/itex] right?

    And is [itex]\mathcal{S}[/itex] really that entire paraboloid or just some part of it?

    Well, the first step is to figure out what curve you are going to integrate over....whaat does Stokes theorem say about that?

    Once you know what curve you are integrating over, you can parameterize the curve and easily determine the limits of integration.
     
  4. Sep 14, 2009 #3
    Yes, S is the surface the paraboloid and yes it is [itex]\int_{\mathcal{S}} (\mathbf{\nabla}\times\textbf{A})\cdot d\textbf{a}[/itex].

    Stokes theorem says that you integrate over the boundary curve, so it would be the circle at the "end" of the paraboloid, with radius 1-z.

    I'm still not sure how to parametrize it. If I used, for A:

    [tex]y=sin(t)[/tex]

    [tex]x=cos(t)[/tex]

    [tex]z=z[/tex]

    With the boundary's parametrization given as:

    [tex]y=(1-z)sin(t)[/tex]

    [tex]x=(1-z)cos(t)[/tex]

    [tex]z=z[/tex]

    I get an answer of zero for the integral. This parametrization is clearly incorrect but I feel as though I was getting somewhere with the cylindrical coordinates. If you think of the paraboloid in terms of a contour plot, at each z height, s is constant (and obviously z is too). So you could have dz = 0 and ds = 0 so d[tex]\phi[/tex] would be your only parameter to integrate over.
     
    Last edited: Sep 14, 2009
  5. Sep 14, 2009 #4

    gabbagabbahey

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    Surely [itex]\mathcal{S}[/itex] isn't the entire parabaloid [itex]z=1-x^2-y^2[/itex]...if it were, [itex]x[/itex] and [itex]y[/itex] would both range from [itex]-\infty[/itex] to [itex]\infty[/itex] and so the radius at the boundary of the surface would be infinite....what is the exact wording of the original question?
     
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