Verify Divergence Theorem for Q with G(x,y,z) in $\Re^3$

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Homework Help Overview

The discussion revolves around verifying the divergence theorem for a vector field G defined in the context of a unit cube Q in \(\mathbb{R}^3\). The divergence theorem relates the volume integral of the divergence of a vector field over a region to the surface integral of the field over the boundary of that region.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss evaluating the surface integral by considering contributions from each side of the cube. There is a focus on understanding how to compute the integral for one specific face of the cube and how to combine results from all faces.

Discussion Status

The discussion is progressing with participants clarifying their understanding of the surface integral and confirming the additive property of integrals. There is an ongoing exploration of how to approach the problem without reaching a final conclusion.

Contextual Notes

Participants mention different notations for the boundary of the region, indicating potential variations in terminology or conventions used in different texts.

Tony11235
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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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Well, you've got six sides on your surface, right?
Treat the contribution from each side separately.
 
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?
 
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]?
 
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].
 
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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