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Math Help - [SOLVED] Unsure - Surface Integral Q?

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    [SOLVED] Unsure - Surface Integral Q?

    Hey everyone!

    I have the following question:
    In the following cases you are given a 3D velocity vector field. This represents the velocity of a fluid at the position (x,y,z). Find the volume of the fluid that is flowing out of the unit cube per second. The density of the fluid is 1g/m^3. Use the simplest possible method to solve when v = 2xyi + 2y^2j - 3yzk.

    I am unsure about what method to take. We are currently learning about vector fields, Green's Theorem, etc. So I'm assuming the method is somehow similar to these. Would it be a surface integral and if so how would I go about solving it?
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    Quote Originally Posted by falconed View Post
    Hey everyone!

    I have the following question:
    In the following cases you are given a 3D velocity vector field. This represents the velocity of a fluid at the position (x,y,z). Find the volume of the fluid that is flowing out of the unit cube per second. The density of the fluid is 1g/m^3. Use the simplest possible method to solve when v = 2xyi + 2y^2j - 3yzk.

    I am unsure about what method to take. We are currently learning about vector fields, Green's Theorem, etc. So I'm assuming the method is somehow similar to these. Would it be a surface integral and if so how would I go about solving it?

    The best method for this would be to use the divergence (Guass's) Theorem. Otherwise you will need to evaluate 6 surface integrals

    \iint_S F \cdot dS=\iiint_V \nabla F dV

    \iiint (2y+4y-3y)dV=3\int_{0}^{1}\int_{0}^{1}\int_{0}^{1} ydxdydz=\frac{3}{2}
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    Quote Originally Posted by TheEmptySet View Post
    The best method for this would be to use the divergence (Guass's) Theorem. Otherwise you will need to evaluate 6 surface integrals

    \iint_S F \cdot dS=\iiint_V \nabla F dV

    \iiint (2y+4y-3y)dV=3\int_{0}^{1}\int_{0}^{1}\int_{0}^{1} ydxdydz=\frac{3}{2}
    Thanks for that! I did it and also got \frac{3}{2}. However, does the density of the fluid ( 1g/m^3) ever become involved into the solution?
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    Quote Originally Posted by falconed View Post
    Thanks for that! I did it and also got \frac{3}{2}. However, does the density of the fluid ( 1g/m^3) ever become involved into the solution?
    This is the flux across the surface i.e how much stuff (Volume)is moving across the surface per second

    \frac{3}{2}\frac{m^3}{s}

    Now if you multiply this by the density

    \frac{3}{2}\frac{m^3}{s}\cdot \frac{1 g}{m^3}=\frac{3}{2}\frac{g}{s}

    This is how many grams are moving across the surface per second.
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