I'm revising for an exam and there will be a question on Gauss' theorem. The question is optional but I am stumped by how to do it and so thought I would post it here. Any help would be greatly appreciated. So the question is:
Verify this theorem for the vector field F = yj + k and for the volume V enclosed by the four planes x = 0, y = 0, z = 0 and 3x + y + 2z = 6.
What precedes the bit in bold is just the definition of Gauss' theorem so I omitted it. The bit I am having a problem with is working out the limits of integration from the information given. I know my lower bounds are x = 0, y = 0 and z = 0, but I don't know how to figure out what my upperbounds are. Thanks.
For the volume integral, your upper bounds will depend on the order of integration. If you integrate $x$ first, then $y$, then $z$, then you should have $$\int_0^3\int_0^{6 - 2z}\int_0^{(6 - y - 2z)/3} \dots ~dx~dy~dz$$
For the surface integral, of course it depends on how you want to parametrize them. The three faces with one of the coordinates $0$ can be parametrized by the other two coordinates. For example, the face $z = 0$ will yield an integration $$\int_0^2\int_0^{6-3x}\dots~dy~dx$$.
The fourth face, lying in the plane $3x + y + 2z = 6$, can be parametrized by any pair of the coordinates, your choice as to which. In this case, it will have the same limits of integration as the other face having the same two coordinates. However, the integrand will be different.