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Divergence Theorm example for 28 Nov 12:00

 
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Nov27-12, 08:01 AM   #1
 

Divergence Theorm example for 28 Nov 12:00


1. The problem statement, all variables and given/known data

Let S be a smooth surface enclosing the volume V, and let [itex]\vec{n}[/itex] to be the unit outward normal. Using the Divergence Theorm show that:


∫∫ x [itex]\vec{r}[/itex] ° [itex]\vec{n}[/itex] dS = 4 * ∫∫∫ x dV,

where [itex]\vec{r}[/itex]=(x,y,z)

2. Relevant equations

Divergence theorm

http://www.math.oregonstate.edu/home...rg/diverg.html

3. The attempt at a solution

I tried to change the form of the those two equations to the form stated in divergence theorm and then to compare the u (or F as in link above), but the u (F) on the left hand side is never equal to this on the right.
 
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Nov27-12, 08:10 AM   #2
 
Your link is broken. But let's suppose the divergence theorem says [tex]\iint (\vec{F} \cdot \vec{n}) dS = \iiint \nabla \cdot \vec{F} dV. [/tex] Now you are given [itex] \vec{F}= x \vec{r} [/itex]. Can you calculate [itex] \nabla \cdot \vec{F} [/itex]?
 
Nov27-12, 08:18 AM   #3
 
[itex]\nabla[/itex] [itex]\cdot[/itex] [itex]\vec{F}[/itex] = (d/dx, d/dy, d/dz) [itex]\cdot[/itex] (x^2,xy,xz) = 2x+x+x=4x
 
Nov27-12, 08:41 AM   #4
 

Divergence Theorm example for 28 Nov 12:00


Quote by debian View Post
[itex]\nabla[/itex] [itex]\cdot[/itex] [itex]\vec{F}[/itex] = (d/dx, d/dy, d/dz) [itex]\cdot[/itex] (x^2,xy,xz) = 2x+x+x=4x
Good job. :)
 
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