[SOLVED] Unsure - Surface Integral Q?

**falconed** · March 27th 2009, 05:22 PM

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 = $2xy$ i + $2y^2$ j - $3yz$ k.

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?

**TheEmptySet** · March 28th 2009, 08:46 AM

Originally Posted by falconed

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 = $2xy$ i + $2y^2$ j - $3yz$ k.

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

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

$\iiint (2y+4y-3y)dV=3\int_{0}^{1}\int_{0}^{1}\int_{0}^{1} ydxdydz=\frac{3}{2}$

**falconed** · March 28th 2009, 04:11 PM

Originally Posted by TheEmptySet

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?

**TheEmptySet** · March 28th 2009, 04:19 PM

Originally Posted by falconed

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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