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Math Help - Surface integral

  1. #1
    MHF Contributor Bruno J.'s Avatar
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    Surface integral

    Let \vec F(x,y,z)=(P(x,y,z),Q(x,y,z),R(x,y,z)) be a vector field on \mathbb R^3 with continuous partial derivatives, and S a smooth orientable surface. Show that

    \iint_S \vec F \cdot \vec{dS} = \iint_S Pdydz + Qdzdx+Rdxdy.


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  2. #2
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by Bruno J. View Post
    Let \vec F(x,y,z)=(P(x,y,z),Q(x,y,z),R(x,y,z)) be a vector field on \mathbb R^3 with continuous partial derivatives, and S a smooth orientable surface. Show that

    \iint_S \vec F \cdot \vec{dS} = \iint_S Pdydz + Qdzdx+Rdxdy.


    I'd go to the definition of the divergence theorem.

     \iint_S \vec F \cdot \vec{dS} = \iiint_V div F dV

     \iiint_V ( \frac{ \partial P }{ \partial x } + \frac{ \partial Q }{ \partial y } + \frac{ \partial R }{ \partial z })dxdydz

    Let's multiply through and examine the first component

     \iiint_V ( \frac{ \partial P }{ \partial x } ) dxdydz

    Note that,

     \int \frac{ \partial P }{ \partial x } dx

    Represents the integral of the derivative of P with respect to x. Which means, we take the derivative and then integrate back to our original function. This means,

     \int \frac{ \partial P }{ \partial x } dx = P

    Thus,

     \iiint_V ( \frac{ \partial P }{ \partial x } ) dxdydz = \iint Pdydz
    Last edited by AllanCuz; May 22nd 2010 at 09:32 AM.
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  3. #3
    MHF Contributor Bruno J.'s Avatar
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    The surface doesn't need to be closed, in which case you cannot use the divergence theorem!
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