Divergence Theorem

  • Thread starter Tony11235
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  • #1
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Let Q denote the unit cube in [tex]\Re^3[/tex] (that is the unite cube with 0<x,y,z<1). Let G(x,y,z) = (y, xe^z+3y, y^3*sinx). Verify the validity of the divergence theorem.

[tex] \int_{Q} \bigtriangledown} \cdot G dxdydz = \int_{\partial Q} G \cdot n dS [/tex]

I am not sure how to evaluate the right side. Any help would be good.
 

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  • #2
arildno
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Well, you've got six sides on your surface, right?
Treat the contribution from each side separately.
 
  • #3
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What would say..the first integral of the six look like? I just need one example.
 
  • #4
arildno
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Okay, let's look at the side x=1, 0<=y,z<=1.
Here, the normal vector is +i.
G(1,y,z)=(y,e^z+3y,y^3sinx)
Forming the dot product between G and the normal vector yields the integrand "y".
This is easy to integrate over the y,z-square (yielding 1/2 in contribution)

Okay?
 
  • #5
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Ok that makes sense. So in the end I add the value of each of the six integrals for the total value of the original integral, [tex]\int_{\partial Q} G \cdot n dS [/tex]?
 
  • #6
Hurkyl
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By the way, is your book actually using [itex]\partial\Omega[/itex]? I'm more used to seeing [itex]d\Omega[/itex] or [itex]\delta\Omega[/itex].
 
  • #7
arildno
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Tony11235 said:
Ok that makes sense. So in the end I add the value of each of the six integrals for the total value of the original integral, [tex]\int_{\partial Q} G \cdot n dS [/tex]?
That's correct.
Remember that the integral operation has the additive property; this entails that you may split up the surface in an arbitrary number of sub-surfaces, calculate the individual contributions and then add the individual contributions together to gain the correct value.
 

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