Very simple question about Stokes theorem / divergence theorem

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

• That's not very relevant actually – Dorian Jul 9 at 19:09
• 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. – Ted Shifrin Jul 9 at 19:59
• Oh, and your volume is missing a factor of $3$, isn't it? – Ted Shifrin Jul 9 at 20:08
• Thank you, that link has been very useful – 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, $$= 81 \int^{2 \pi}_0 \int^{\pi}_0 \bigg( \cos^2 \theta \sin^3 \phi + \dfrac{2}{3} \sin^2 \theta \sin^3 \phi \bigg) d \theta d \phi = 180 \pi$$
• 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 $r^2$. – Ted Shifrin Jul 10 at 14:02