TITLE: Counter example for the divergence theorem (i.e I am not understanding something)?
QUESTION [1 upvotes]: The divergence theorem states:
$\int\int_{\partial W}F\cdot dS = \int\int\int_W div(F) dV$
Assume $F=\langle 7x, 3z, 8y \rangle$
Assume the volume is the cylinder described by:
$x^2 + y^2 \leq 1$, $0\leq z \leq 6$
Computing the left hand derivative yields: 
$\int\int_{\partial W}F\cdot dS = 42\pi$ (this is the correct answer to the problem)
Knowing thus that the answer is $42\pi$ we do:
$div(F) = 7+0+0 = 7$
And so the right hand side gives:
$ \int\int\int_W div(F) dV = \int^{2\pi}_0\int^1_0\int^6_0 7 dt$ $dr$ $d\theta = 84\pi$
The cylinder has radius 1 and height 6, trivially. So the bounds should not be the problem, and the divergence seems easy enough. So what is the problem here?

REPLY [1 votes]: You need an r into the integral.
$$ \int^{2\pi}_0\int^1_0\int^6_0 7 r dt dr d\theta = 42\pi$$
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