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Math Help - Theorem of divergence and Stoke

  1. #1
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    Theorem of divergence and Stoke

    Where can I use each theorem ?
    Could you show me an example?

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    One major example of the Divergence Theorem is Gauss's Law, which is a flux integral:

    \oint_S \bold{E}.\bold{dA} = \frac{Q_{enclosed}}{\epsilon_0}

    More information on Divergence Theorem here: Divergence theorem - Wikipedia, the free encyclopedia
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  3. #3
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    Quote Originally Posted by Aryth View Post
    One major example of the Divergence Theorem is Gauss's Law, which is a flux integral:

    \oint_S \bold{E}.\bold{dA} = \frac{Q_{enclosed}}{\epsilon_0}

    More information on Divergence Theorem here: Divergence theorem - Wikipedia, the free encyclopedia
    Ok. Thank you
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  4. #4
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    Stokes I always use to compute flux ? And the divergence theorem ?

    example

    Determine the flux vector field: F(x,y,z) = xi +yj + zk through the paraboloid z = x^2+y^2 below the plane z = 1 and vector normal to the surface points to the external side of the paraboloid

    I can use Stokes or divergence theorem ?
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  5. #5
    MHF Contributor Amer's Avatar
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    the flux equal

    flux=\int\int F.n ds

    so how you can use Stokes's theorem for flux is there anti curl because Stoke's

    \oint F.dr=\int\int \nabla\times F .n dS

    since \nabla\times F is the curl of F so if there is an anti for curl you can use Stoke's for flux
    but the divergent theorem is

    \int\int F.n dS = \int\int\int div F dv

    divF =\nabla F is the gradient

    Stoke's theorem change from surface integral to linear integral but divergence theorem change from surface integral to volume

    Green's is a special case from Stoke's

    that what I think is correct is there any mistake about I said ...
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  6. #6
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    in the example that I showed, if I use curl (equal 0) the flux will be equal to 0?
    If I use the divergence theorem ?
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  7. #7
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Apprentice123 View Post
    in the example that I showed, if I use curl (equal 0) the flux will be equal to 0?
    If I use the divergence theorem ?
    I think you misunderstand me the flux equal

    \int\int F.n dS \ne \int\int \nabla\times F .n dS

    I need the function that

    F(x,y,z) = \nabla\times G(x,y,z) if I can find G I can apply stoke's

    \int\int F.n dS = \int\int \nabla\times G.n.dS = \oint G.dr
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  8. #8
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    Quote Originally Posted by Amer View Post
    I think you misunderstand me the flux equal

    \int\int F.n dS \ne \int\int \nabla\times F .n dS

    I need the function that

    F(x,y,z) = \nabla\times G(x,y,z) if I can find G I can apply stoke's

    \int\int F.n dS = \int\int \nabla\times G.n.dS = \oint G.dr
    It is possible use Stokes in the previous example ?
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  9. #9
    MHF Contributor Amer's Avatar
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    ok show me you should find G such that

     F = \nabla\times G

    you can use Sotke's for flux if they give you a vector like this

    \nabla\times V(x,y,z)

    so for the flux

    \int\int \nabla\times V(x,y,z).n dS = \oint V .dr

    you can't use divergence if the surface is not closed if it is closed the integral above equal zero because

    \int\int \nabla\times V .n .dS = \int\int\int div(\nabla\times V) dv = 0

    because

    div(\nabla\times V )=0

    div for the curl equal zero
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  10. #10
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    Quote Originally Posted by Amer View Post
    ok show me you should find G such that

     F = \nabla\times G

    you can use Sotke's for flux if they give you a vector like this

    \nabla\times V(x,y,z)

    so for the flux

    \int\int \nabla\times V(x,y,z).n dS = \oint V .dr

    you can't use divergence if the surface is not closed if it is closed the integral above equal zero because

    \int\int \nabla\times V .n .dS = \int\int\int div(\nabla\times V) dv = 0

    because

    div(\nabla\times V )=0

    div for the curl equal zero

    F(x,y,z) = xi+yj+zk

    xi+yj+zk = \nabla X G(x,y,z)

    How do I find G?
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  11. #11
    MHF Contributor Amer's Avatar
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    ok

    curl G = F do you know the curl l determinate l

    l....... i...... j...... k............ l

    l..... d/dx.. d/dy...... d/dz... l = xi + yj + zk

    l ...... f...... h........... g..... l


    i(\frac{\partial g}{\partial y} -\frac{\partial h}{\partial z} ) -j(\frac{\partial g}{\partial x} -\frac{\partial f }{\partial z})+k(\frac{\partial h}{\partial x} -\frac{\partial f }{\partial y})=xi+yj+zk   <br />

    (\frac{\partial g}{\partial y} -\frac{\partial h}{\partial z} ) = x

    (\frac{\partial g}{\partial x} -\frac{\partial f }{\partial z}) = y

    (\frac{\partial h}{\partial x} -\frac{\partial f }{\partial y}) = z

    first you should find the derivative with respect to x,y,z then you should find the f ,h, g after you find them but them like this

    F(x,y,z)=f(x,y,z)i + g(x,y,z)j + h(x,y,z)k

    I will try to find them
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  12. #12
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    I think the flux is 0

    Paraboloid below the plane z = 1 is a closed surface

    F(x,y,z) = Mi + Nj + Pk

    [ d(N)/dx - d(M)/dy ]dxdy = 0

    [ d(P)/dx - d(M)/dz ]dxdz = 0

    [ d(P)/dy - d(N)/dz ]dydz = 0
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  13. #13
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Apprentice123 View Post
    Stokes I always use to compute flux ? And the divergence theorem ?

    example

    Determine the flux vector field: F(x,y,z) = xi +yj + zk through the paraboloid z = x^2+y^2 below the plane z = 1 and vector normal to the surface points to the external side of the paraboloid

    I can use Stokes or divergence theorem ?
    no

    the flux
    I will use divergence

    \int\int F.n dS=\int\int\int div(F) dv

    \int\int\int (\frac{\partial}{\partial x}i+\frac{\partial}{\partial x}j+\frac{\partial}{\partial x}k)(xi+yj+zk) dv

    \int\int\int \left(\frac{\partial x}{\partial x}+\frac{\partial y}{\partial y}+\frac{\partial z}{\partial z}\right) dv=\int\int\int (1+1+1) dv=3\int\int\int dv=3V V volume of paraboloid
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  14. #14
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    Quote Originally Posted by Amer View Post
    no

    the flux
    I will use divergence

    \int\int F.n dS=\int\int\int div(F) dv

    \int\int\int (\frac{\partial}{\partial x}i+\frac{\partial}{\partial x}j+\frac{\partial}{\partial x}k)(xi+yj+zk) dv

    \int\int\int \left(\frac{\partial x}{\partial x}+\frac{\partial y}{\partial y}+\frac{\partial z}{\partial z}\right) dv=\int\int\int (1+1+1) dv=3\int\int\int dv=3V V volume of paraboloid
    Yes I thought about it too. But no important that the paraboloid is below the plane z = 1?

    You know the volume the any paraboloid?

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  15. #15
    MHF Contributor Amer's Avatar
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