View Full Version : Divergence Theorem
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.
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.
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 ?
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.
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