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?



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*}

  • $\begingroup$ 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, $2\cdot\pi$ $\endgroup$ – Relure Jun 14 '17 at 18:56
  • $\begingroup$ Those are the right questions. Let me edit the answer. $\endgroup$ – Matthew Leingang Jun 14 '17 at 21:47
  • $\begingroup$ I understand your solution now! Then, the solution I pointed out in my comment is also correct? $\endgroup$ – Relure Jun 14 '17 at 22:25
  • $\begingroup$ @Relure: Yes, yours is right, too. It's just that your $S_0$ and mine are oriented oppositely, so the minus signs cancel. $\endgroup$ – Matthew Leingang Jun 15 '17 at 0:36
  • 1
    $\begingroup$ @Aladdin no, in this case $S$ is part of the boundary of the solid $V$, but it is not the graph of a function. $\endgroup$ – Matthew Leingang Jan 28 '20 at 16:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.