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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?

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    $\begingroup$ The Jacobian for cylindrical coordinates means that inside that last integral you should have $7r$, not $7$. $\endgroup$ – Joppy Dec 11 '17 at 0:00
  • $\begingroup$ shoot, thank you, it's quite confusing that sometimes you need the jacobian and sometimes you don't thank you $\endgroup$ – Makogan Dec 11 '17 at 0:02
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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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