[SOLVED] Unsure - Surface Integral Q?

Printable View

• Mar 27th 2009, 05:22 PM
falconed
[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 = $\displaystyle 2xy$i + $\displaystyle 2y^2$j - $\displaystyle 3yz$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?
• Mar 28th 2009, 08:46 AM
TheEmptySet
Quote:

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 = $\displaystyle 2xy$i + $\displaystyle 2y^2$j - $\displaystyle 3yz$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

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

$\displaystyle \iiint (2y+4y-3y)dV=3\int_{0}^{1}\int_{0}^{1}\int_{0}^{1} ydxdydz=\frac{3}{2}$
• Mar 28th 2009, 04:11 PM
falconed
Quote:

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

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

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

Thanks for that! I did it and also got $\displaystyle \frac{3}{2}$. However, does the density of the fluid ($\displaystyle 1g/m^3$) ever become involved into the solution?
• Mar 28th 2009, 04:19 PM
TheEmptySet
Quote:

Originally Posted by falconed
Thanks for that! I did it and also got $\displaystyle \frac{3}{2}$. However, does the density of the fluid ($\displaystyle 1g/m^3$) 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

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

Now if you multiply this by the density

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

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