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Tony11235
Sep9-05, 10:00 AM
Let Q denote the unit cube in \Re^3 (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.

\int_{Q} \bigtriangledown} \cdot G dxdydz = \int_{\partial Q} G \cdot n dS

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

arildno
Sep9-05, 10:03 AM
Well, you've got six sides on your surface, right?
Treat the contribution from each side separately.

Tony11235
Sep9-05, 10:17 AM
What would say..the first integral of the six look like? I just need one example.

arildno
Sep9-05, 10:21 AM
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?

Tony11235
Sep9-05, 10:40 AM
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, \int_{\partial Q} G \cdot n dS ?

Hurkyl
Sep9-05, 05:19 PM
By the way, is your book actually using \partial\Omega? I'm more used to seeing d\Omega or \delta\Omega.

arildno
Sep10-05, 06:30 AM
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, \int_{\partial Q} G \cdot n dS ?
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.