# Problem finding the flux over a cylinder

I'm doing this exercise:

Let's consider the vector field given by $\mathbf F(x,y,z)=(x+1,y-1,1-2z)$,and the cylinder given by $S=\{x^2+y^2=1\mid\ 0≤z≤1 \}$, with the orientation given by $\mathbf{n}=\displaystyle\frac{1}{\sqrt{x^2+y^2}}(x,y,0)$. Calculate the flux over the surface $S$ integrating the divergence over a situable domain.

My try:

If we calculate the divergence and we use the Gauss theorem, we see that $$\iint_{S}\mathbf F\cdot dS=\iiint_V\operatorname{div}(\mathbf F)\,dV$$ but $\operatorname{div}(\mathbf F)=1+1-2,$ so the flux over any surface is $0$.

Is there something I'm missing?

Thanks.

The surface $S$ is not the entire boundary of $V$. To use the divergence theorem, you have to close up $S$ by adding the top and bottom disks. Let \begin{align*} S_1 &= \left\{(x,y,1)\mid x^2 + y^2 \leq 1 \right\} \\ S_0 &= \left\{(x,y,0)\mid x^2 + y^2 \leq 1 \right\} \\ \end{align*} So as surfaces, $$\partial V = S \cup S_1 \cup S_0$$
Now we need to orient those surfaces. The conventional way to orient a surface that is the graph of a function is “upwards”—that is, the normal direction with positive $z$-component. So $\mathbf{n} = \mathbf{k}$ on $S_0$ and $S_1$. The conventional way to orient a surface that is the boundary of a solid is “outwards.” On $S_1$, upwards and outwards are the same, but on $S_0$, upwards and outwards are opposite. So we say $$\partial V = S + S_1 - S_0$$ as oriented surfaces. Therefore \begin{align*} \iiint_V \operatorname{div}\mathbf{F}\,dV &= \iint_{\partial V} \mathbf{F} \cdot d\mathbf{S} \\&= \iint_{S + S_1 - S_0} \mathbf{F} \cdot d\mathbf{S} \\&= \iint_S \mathbf{F} \cdot d\mathbf{S} + \iint_{S_1} \mathbf{F} \cdot d\mathbf{S} - \iint_{S_0} \mathbf{F} \cdot d\mathbf{S} \end{align*} The three integrals on the right are no longer thought of as attached to $V$. In particular, $S_1$ and $S_0$ are both oriented upwards now. We can solve this for $\iint_S \mathbf{F} \cdot d\mathbf{S}$: $$\iint_S \mathbf{F} \cdot d\mathbf{S} = \iiint_V \operatorname{div}\mathbf{F}\,dV -\iint_{S_1} \mathbf{F} \cdot d\mathbf{S} +\iint_{S_0} \mathbf{F} \cdot d\mathbf{S}$$ As you point out, the first integral on the right-hand side is zero. On the surface $S_1$, $\mathbf{n} = \mathbf{k}$ and $\mathbf{F} = \left<x+1,y-1,-1\right>$. So $$\iint_{S_1} \mathbf{F} \cdot d\mathbf{S} =\iint_{S_1} \left<x+1,y-1,-1\right>\cdot \mathbf{k}\,dS =\iint_{S_1} (-1)\,dS = (-1) \operatorname{Area}(S_1) = -\pi$$ On $S_2$, $\mathbf{F} = \left<x+1,y-1,1\right>$, so $$\iint_{S_0} \mathbf{F} \cdot d\mathbf{S} =\iint_{S_0} 1\,dS = \operatorname{Area}(S_1) = \pi$$ Putting these all together, we have $$\iint_S \mathbf{F} \cdot d\mathbf{S} = 2\pi$$
In this case, it's possible to check the calculation by computing the flux directly. Parametrize the surface $S$ by $$\mathbf{r}(u,v) = \left<\cos u,\sin u, v\right>$$ over $0 \leq u \leq 2\pi$, $0 \leq v \leq 1$. Then $$\mathbf{n} = \left<\cos u,\sin u,0\right>$$ So \begin{align*} \iint_S \mathbf{F} \cdot d\mathbf{S} &= \int_0^{2\pi}\int_0^1 \left<\cos u+1,\sin u-1, v\right>\cdot \left<\cos u,\sin u,0\right>\,dv\,du \\ &= \int_0^{2\pi}\int_0^1 \left(\cos^2 u+\cos u +\sin^2 u-\sin u\right)\,dv\,du \\ &= \int_0^{2\pi} \left(\cos^2 u+\cos u +\sin^2 u-\sin u\right)\,du \\ &= \int_0^{2\pi} \left(1+\cos u -\sin u\right)\,du \\ &= 2\pi + 0 - 0 = 2\pi \end{align*}
• Okay, I have 2 questions now: How do you decide in which direction you take n on $S_1$ and $S_0$? And why did you decide to add the $S_1$ integral and take out the $S_0$ one? If I understood it correctly, we should have: $$\iint_S \mathbf{F} \cdot d\mathbf{S} = \iiint_V \operatorname{div}\mathbf{F}\,dV -\iint_{S_1} \mathbf{F} \cdot d\mathbf{S} -\iint_{S_0} \mathbf{F} \cdot d\mathbf{S}$$ right? On $S_0$, the normal it's $\mathbf{n} = \mathbf{-k}$, and then the integral over $S_0$ is $-\pi$, giving when you add up everything, $2\cdot\pi$ – Relure Jun 14 '17 at 18:56
• @Relure: Yes, yours is right, too. It's just that your $S_0$ and mine are oriented oppositely, so the minus signs cancel. – Matthew Leingang Jun 15 '17 at 0:36
• @Aladdin no, in this case $S$ is part of the boundary of the solid $V$, but it is not the graph of a function. – Matthew Leingang Jan 28 '20 at 16:43