I know these two theorems:

$\iiint_V(\nabla\cdot\mathbf F)dV= \iint_S(\mathbf F\cdot\mathbf n)dS $, with $V\subset\mathbb R^3$ compact, $S=\partial V,\ \mathbf F\in\mathfrak X(\Bbb R^3)$ and $\mathbf n$ is the normal unitary vector field for $S$ (divergence theorem).

$\int_{\partial\Omega}\omega=\int_{\Omega}d\omega$, with $\omega$ a differential form on the compact manifold $\Omega$ (Stokes theorem).

However I'm given this simple exercise:

Find the flux of $\xi=3x\mathbf i+2y\mathbf j\in\mathfrak X(\Bbb R^3)$ through the surface of $S=\{(x \ y\ z)^t:x^2+y^2+z^2=9\}$, oriented with a normal unitary vector field pointing outward $S$.

Now, if I apply the divergence theorem, since div$(\xi)=5$, I have $\iiint_V5dV=5\cdot\frac 4 3\pi\cdot 9=$ $=\frac {20} 3\pi$. However if I try to find the flux with Stokes, I obtain a different result: since $\mathbf n=\frac1 3(x\mathbf i+y\mathbf j+z\mathbf k)$, $\mathbf F\cdot\mathbf n=x^2+\frac 2 3 y^2$, and so $\iint_S(\mathbf F\cdot\mathbf n)dS =\int_S(x^2+\frac 2 3 y^2)(\frac 1 3xdy\wedge dz-\frac 1 3ydx\wedge dz+\frac 1 3zdx\wedge dy)$. Clearly if I derive this differential $2-$form I don't obtain the form $5dx\wedge dy \wedge dz$, so actually this two methods give me two different results. Where am I wrong? Thank you in advance

  • $\begingroup$ That's not very relevant actually $\endgroup$ – Dorian Jul 9 at 19:09
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    $\begingroup$ Did you actually do the integration? You'll get the right answer. But for the "right" way to do the flux computation, see this question. $\endgroup$ – Ted Shifrin Jul 9 at 19:59
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    $\begingroup$ Oh, and your volume is missing a factor of $3$, isn't it? $\endgroup$ – Ted Shifrin Jul 9 at 20:08
  • $\begingroup$ Thank you, that link has been very useful $\endgroup$ – Dorian Jul 10 at 11:52

You've got your volume wrong, as $V = \dfrac{4 \pi}{3} r^3$, which on putting here gives $$ 5 \iiint_V dV = 180 \pi $$ For surface integral, you've got $\mathbf F = 3x \hat i + 2 y \hat j, \mathbf n = \dfrac{1}{3} ( x \hat i + y \hat j + z \hat k)$

Which makes $$\iint_S \mathbf F \cdot \mathbf n dS = \iint_D \bigg( x^2 + \dfrac{2}{3} y^2 \bigg) dA$$

Which in spherical coordinates, $$ = 81 \int^{2 \pi}_0 \int^{\pi}_0 \bigg( \cos^2 \theta \sin^3 \phi + \dfrac{2}{3} \sin^2 \theta \sin^3 \phi \bigg) d \theta d \phi = 180 \pi $$

  • $\begingroup$ Thank you very much :) $\endgroup$ – Dorian Jul 10 at 11:51
  • $\begingroup$ Easier still: Use symmetry to deduce that the integral of $x^2$ and the integral of $y^2$ are each $1/3$ the integral of $r^2$. $\endgroup$ – Ted Shifrin Jul 10 at 14:02

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