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Math Help - divergence theorem

  1. #1
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    divergence theorem

    Can I use the divergence theorem to calculate the following:

     \vec{G} = -x^3\vec{i}  - y^3\vec{j}  - z^3\vec{k}

    Find the flux of the vector field out of the closed surface S. S is the boundary of the solid region W. W is the upper hemisphere of radius 3 centered at the origin.

    well the divG I got is -3x^2 -3y^2 -3z^2, which is  -3(x^2 + y^2 + z^2) and this translates to  -3r^2 am I right? do I just then multiply this by the volume, which is  18\pi in this case?
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  2. #2
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    Quote Originally Posted by TheRekz View Post
    Can I use the divergence theorem to calculate the following:

     \vec{G} = -x^3\vec{i}  - y^3\vec{j}  - z^3\vec{k}

    Find the flux of the vector field out of the closed surface S. S is the boundary of the solid region W. W is the upper hemisphere of radius 3 centered at the origin.

    well the divG I got is -3x^2 -3y^2 -3z^2, which is  -3(x^2 + y^2 + z^2) and this translates to  -3r^2 am I right? do I just then multiply this by the volume, which is  18\pi in this case?
    You do know what the divergence theorem says, don't you?

    \int\int \vec{G}\cdot d\vec{S}= \int\int\int \nabla\cdot \vec{G} dV
    Since \nabla\cdot \vec{G} is not constant in that solid, no, you do not "multiply by the volume", you integrate.
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    Quote Originally Posted by HallsofIvy View Post
    You do know what the divergence theorem says, don't you?

    \int\int \vec{G}\cdot d\vec{S}= \int\int\int \nabla\cdot \vec{G} dV
    Since \nabla\cdot \vec{G} is not constant in that solid, no, you do not "multiply by the volume", you integrate.
    So I would have to translate that to a spherical coordinate first and integrate it?
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    Quote Originally Posted by TheRekz View Post
    So I would have to translate that to a spherical coordinate first and integrate it?
    Don't ask. Do! Then post your solution for comment.
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    Quote Originally Posted by HallsofIvy View Post
    You do know what the divergence theorem says, don't you?

    \int\int \vec{G}\cdot d\vec{S}= \int\int\int \nabla\cdot \vec{G} dV
    Since \nabla\cdot \vec{G} is not constant in that solid, no, you do not "multiply by the volume", you integrate.
     -3 \int_0^{2\pi}\int_0^{\pi/2}\int_0^3 \rho^2 sin \phi d\rho d\phi d\theta

    and the final result I have is -54pi... and it doesn't match with what the computer says
    Last edited by TheRekz; April 28th 2009 at 09:21 AM.
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    Quote Originally Posted by TheRekz View Post
     -3 \int_0^{2\pi}\int_0^{\pi/2}\int_0^3 \rho^2 sin \phi d\rho d\phi d\theta

    and the final result I have is -54pi... and it doesn't match with what the computer says
    The answer is given by  -3 \int_0^{2\pi}\int_0^{\pi/2}\int_0^3 \rho^{\color{red}4} sin \phi d\rho d\phi d\theta .
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    may I ask why it's to the power of 4?
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    Quote Originally Posted by TheRekz View Post
    may I ask why it's to the power of 4?
    Instead of posting so soon (within a minute of my post!) you should have spent time thinking about what a posted.

    Do you know what dV is in spherical coordinates? Do you recall that the thing you're integrating (the divegrence) is -3r^2? Think carefully about these two questions.
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    thanks
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