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Math Help - flux integral help

  1. #1
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    flux integral help

    Hey,

    Let f(x,y) = ln(x^2 +y^2)

    Let C be the circle x^2 + y^2 =a^2 with a^2 \ne 0

    and evaluate the flux integral:

    \oint_C\nabla f \cdot n ds

    I can't seem to use Greens Theorem on this particular question and don't know why.

    Second part:
    Now let K be an arbitrary simple closed curve in the plane that does not pass through
    (0,0). Use Green's theorem to show that:

    \oint_K\nabla f \cdot n ds
    has, only, two possible values depending on whether (0,0) lies inside or outside of K.
    I really need help on this one.

    Can someone help me with this?

    Thanks
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  2. #2
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    Quote Originally Posted by Nguyen View Post
    Hey,

    Let f(x,y) = ln(x^2 +y^2)

    Let C be the circle x^2 + y^2 =a^2 with a^2 \ne 0

    and evaluate the flux integral:

    \oint_C\nabla f \cdot n ds

    I can't seem to use Greens Theorem on this particular question and don't know why.

    Second part:
    Now let K be an arbitrary simple closed curve in the plane that does not pass through
    (0,0). Use Green's theorem to show that:

    \oint_K\nabla f \cdot n ds
    has, only, two possible values depending on whether (0,0) lies inside or outside of K.
    I really need help on this one.

    Can someone help me with this?

    Thanks
    Note that

    \nabla f(x,y)=\frac{2x}{x^2+y^2}\mathbf{i}+\frac{2y}{x^2+  y^2}\mathbf{j}

    \mathbf{n}=\frac{1}{\sqrt{x^2+y^2}}\left( x\mathbf{i}+ y\mathbf{j}\right)

    \nabla f \cdot \mathbf{n}=\frac{2}{\sqrt{x^2+y^2}}

    Now we just need to compute

    \int \frac{2}{\sqrt{x^2+y^2}}ds

    x=a\cos(t) \quad y=a\sin(t)

    \int_{0}^{2\pi}\frac{2}{a}(adt)=2\int_{0}^{2\pi}dt  =4\pi

    for part 2 if the curve does not contain the origin Greens theorem (Stokes theorem)applies and

    \nabla \times \nabla f=0

    so the integral is zero if it does consider the new domain with a circle of radius a cut out around the origin the greens theorem applies on the domain with the hole and

    \int_{K}\nabla f \cdot \mathbf{n}ds = \int_{D}\nabla f \cdot \mathbf{n}ds + \lim_{a \to 0} \int_{C}\nabla f \cdot \mathbf{n}ds

    By Greens theorem again we get

    0 = \int_{D}\nabla f \cdot \mathbf{n}ds - \lim_{a \to 0}  \int_{C}\nabla f \cdot \mathbf{n}ds

     \int_{D}\nabla f \cdot \mathbf{n}ds = \lim_{a \to 0}  \int_{C}\nabla f \cdot \mathbf{n}ds =4\pi

    The last step comes from part 1 as the integral was independent of a.
    Last edited by TheEmptySet; May 8th 2011 at 07:57 AM. Reason: missing limit
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