r19ecua
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Homework Statement
Suppose the one-dimensional field A = Kx * ax exists in a region. Illustrate the validity of the Gaussian theorem by evaluating its volume and surface integrals inside and on the rectangular parallelepiped bounded by the surfaces: x=1,x=4,y=2,y=-2,z=0 and z=3, for a given A.
Homework Equations
(Right) [itex]\int_0^3[/itex][itex]\int_1^4x[/itex]dxdz ay + (left) [itex]\int_0^3[/itex][itex]\int_1^4x[/itex]dxdz -(ay) + (top) [itex]\int_{-2}^2[/itex][itex]\int_1^4x[/itex]dxdy az + (bottom) [itex]\int_{-2}^2[/itex][itex]\int_1^4x[/itex]dxdy -(az) + (front) [itex]\int_0^3[/itex][itex]\int_{-2}^2[/itex]dydz (ax) + (back) [itex]\int_0^3[/itex][itex]\int_{-2}^2[/itex]dydz -(ax)
Direction on the left is applied to the integral on its right.
The Attempt at a Solution
For the Right side
[itex]\int_0^3[/itex][itex]\int_1^4x[/itex]dxdz ay
My answer to this integral is 45/2
For the left side
[itex]\int_0^3[/itex][itex]\int_1^4x[/itex]dxdz -(ay)
My answer to this integral is -45/2
For the top side
[itex]\int_{-2}^2[/itex][itex]\int_1^4x[/itex]dxdy az
My answer is 30
For the bottom
[itex]\int_{-2}^2[/itex][itex]\int_1^4x[/itex]dxdy -(az)
-30
For the front
[itex]\int_0^3[/itex][itex]\int_{-2}^2[/itex]dydz (ax)
36
For the back
[itex]\int_0^3[/itex][itex]\int_{-2}^2[/itex]dydz -(ax)
-36
When I add these up, I get zero... however, when I use the divergence theorem I get 36.
This answer is suppose to equal the answer I get via the divergence theorem formula. I'm confused :(