Is the Divergence Theorem Valid for a Specific Vector Field on a Cube?

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SUMMARY

The divergence theorem is validated for the vector field F(r) = <3x, xy, 2xz> on a cube defined by the planes x=0, x=1, y=0, y=1, z=0, and z=1. The divergence of F is calculated as div(F) = 3 + 3x, leading to a volume integral of 4.5. To evaluate the flux through the cube's surfaces, the correct parametrization for the top face is r(x,y,z) = (x, y, 1), which is essential for accurate normal vector calculations. The initial approach to calculate the flux was incorrect due to the misrepresentation of the surface vector.

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Homework Statement


verify that the divergence theorem in 3-d is true for the vector field F(r)=<3x,xy,2xz>
on the cube bounded by the planes x=0 x=1 y=0 y=1 z=0 z=1


Homework Equations





The Attempt at a Solution



so fristly div(F)=d/dx(3x)+d/dy(xy)+d/dz(2xz)=3+3x
[tex]\int[/tex][tex]\int[/tex][tex]\int[/tex] 3+3xdxdydz=4.5

now i need to evaluate flux through each faces of the cube separately so i was just wondering if i am doing this write say i would want to evaluate the top surface of the cube
then i would have to parametrize it so would the following be corret
r(x,y,z)=(3x,xy,1)
dr/dx=(3,y,0)
dr/dy=(0,x,0)
(dr/dx) X (dr/dy) = (0,0,3x)
r(x,y,z).((dr/dx) X(dr/dy))= (3x,xy,1).(0,0,3x) = 3x
[tex]\int[/tex][tex]\int[/tex] 3x dydx
=3/2

and i have to do the same for all other five surfaces so is this the correct way?
 
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You're not using the correct r when calculating the normal. You want to describe the surface, not the vector field. For the top face, it would be r=(x, y, 1).
 
ok thanks
 

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