Let $\mathbf{F} = (3y, -xz, yz^2)$, and let $S=\{(x,y,z): z=\frac{1}{2}(x^2+y^2), z\leq 2\}$. Find $\iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S}$.

Firstly, I know I can compute this quite easily using Stokes' theorem -- my question concerns only the divergence theorem. Using Stokes' theorem, I obtain an answer of $-20\pi$.

Now below is my working relating to the divergence theorem.

If I let $\Sigma$ be the disc $x^2+y^2\leq 4$ on the plane $z=2$, then $S \cup \Sigma$ is the boundary of $\Omega = \{(x,y,z) : \frac{1}{2}(x^2+y^2)\leq z\leq 2\}$. So by the divergence theorem, $$ \iiint_\Omega \nabla \cdot \mathbf{F} dV = \iint_S \mathbf{F}\cdot d\mathbf{S} + \iint_\Sigma \mathbf{F}\cdot d\mathbf{S}. $$ For the region $\Sigma$, the unit normal is $(0,0,1)$, so we can calculate the surface integral as \begin{align*} \iint_\Sigma \mathbf{F}\cdot d\mathbf{S} &= \iint_{\Sigma} \mathbf{F}\cdot (0,0,1) \,dA \\ &= \iint_{x^2+y^2\leq 4 \\ z=2} yz^2 dA \\ &= 4 \iint_{x^2+y^2\leq 4} y dA \\ &= 4 \int_0 ^{2\pi} \int_0^2 (r\sin\theta) drd\theta \\ &= 4 \int_0 ^{2\pi} 2\sin \theta = 0. \end{align*} Also, $\nabla \cdot \mathbf{F} = (0,0,2zy)$, so \begin{align*} \iiint_\Omega 2zy dV &= 2\int_0^2 \iint _{x^2+y^2\leq 2z} yz\,dV \\ &=2\int_0^2\int_0^{2\pi} \int_0^\sqrt{2z} (r\sin\theta z) dr d\theta dz \end{align*} but this will still be $0$ because of the presence of $\sin \theta$. This would imply that $\iint_S \mathbf{F}\cdot d\mathbf{S}=0$.

My question is: what went wrong with the second computation? I think it might be the way I handled the volume integral $\iiint_\Omega$, but I'm not quite sure. Have I misunderstood the divergence theorem? Any guidance would be very much appreciated.

  • $\begingroup$ The question is poorly written. What is the surface &S$ and hoe $\endgroup$ Commented May 16, 2022 at 4:17
  • $\begingroup$ @TedShifrin I'm not sure what you mean by 'hoe'. The surface as given to me is $S=\{(x,y,z): z=\frac{1}{2}(x^2+y^2), z\leq 2\}$ as I originally tried to state (I'll edit to make it clearer). How should I write my question better? $\endgroup$ Commented May 16, 2022 at 4:56
  • $\begingroup$ @user1057909 The question says "find $\iint \nabla \times F dS$" but you're using divergence theorem which give $\iint F\cdot dS$, are you applying the divergence theorem for $F=\nabla\times F$? $\endgroup$
    – A. P.
    Commented May 16, 2022 at 5:12
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    $\begingroup$ It seems to me you've correctly solved the wrong problem: your work is correct for finding $\iint \vec{F} \cdot dS$ but the question asking for $\iint \nabla \times \vec{F} \cdot dS.$ (fuller answer inbound) $\endgroup$ Commented May 16, 2022 at 5:15
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    $\begingroup$ @EventHorizon Oh, that's exactly right. I've figured it out now, thanks!! $\endgroup$ Commented May 16, 2022 at 5:33

2 Answers 2


I've figured out my mistake, thanks to @Event Horizon. My impression from the Help Center page is that I shouldn't delete my question, so I'll outline what went wrong:

I should've applied the divergence theorem to $\nabla \times \mathbf{F}$ instead, so that the statement of the theorem becomes $$ \iiint_\Omega \nabla \cdot (\nabla \times \mathbf{F}) = \iint_S \nabla \times \mathbf{F} \cdot d\mathbf{S} + \iint_\Sigma \nabla \times \mathbf{F}\cdot d\mathbf{S}. $$

The left is $0$ by since the divergence of a curl is $0$. From $\nabla \times \mathbf{F} = (z^2+x,0,-z-3)$, the correct computation is $$ \iint_\Sigma \nabla \times \mathbf{F}\cdot d\mathbf{S} = \iint_{x^2+y^2\leq 4 \\ z=2} (-z-3 )dA = \iint-5dA = -20\pi$$ which matches what I got using Stokes' theorem (up to sign - but this just depends on the orientation of $S$).

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    $\begingroup$ Good job. Regarding your sign issues I think it's worth noting that with the divergence theorem the positive orientation for the closed surface is outward, which would go down through the parabola, versus when you used Stoke's theorem I'll wager you opted to go counterclockwise around the boundary (typically the default) which would give the upward orientation. This explains the sign discrepancy between the two methods. $\endgroup$ Commented May 16, 2022 at 5:54

Just for reinforcement, I thought it would be good to show that your work is correct for $\iint_S \vec F \cdot d\vec S$, and we can show this by working directly:

$$\vec r (x, y) = \langle x, y, \frac12 (x^2 + y^2)\rangle$$ $$\vec n = \vec r_x \times \vec r_y = \langle 1, 0, x\rangle \times \langle 0, 1, y\rangle = \langle -x, -y, 1\rangle$$ $$\iint_S \vec F \cdot d\vec S = \iint_\Sigma \langle 3y, -\frac12 x (x^2 + y^2), \frac14 y(x^2 + y^2)^2\rangle \cdot \frac1{x^2 + y^2+1}\langle-x, -y, 1\rangle dA = \iint_\Sigma \frac1{x^2 + y^2 + 1} \left(-3xy + \frac12xy(x^2+y^2)+\frac14 y(x^2+y^2)^2 \right) dA$$

where $\Sigma$ is defined as the same disk you used.

At first this looks like a bit of a mess, but note that the integrand is odd in $y$ and that the region of integration is symmetric across the $x$-axis. Letting our messy integrand be $f,$ this means that $f(x, -y) = -f(x,y)$ and that if a point $(x,y)$ is in $\Sigma$ then so is $(x, -y).$ (both of which are trivial here)

We can use this by defining $R_1 = \{(x, y) \in \Sigma : y \geq 0\}$ and $R_2 = \{(x, y) \in \Sigma : y \leq 0\},$ noting that also $R_2 = \{(x, -y) : (x, y) \in R_1\}.$ So,

$$\iint_\Sigma f(x,y) dA = \iint_{R_1} f(x,y) dA + \iint_{R_2} f(x,y) dA = \iint_{R_1} f(x,y) dA + \iint_{R_1} f(x,-y) dA = \iint_{R_1} f(x,y) + f(x, -y) dA = \iint_{R_1} 0 dA = \boxed{0}$$

as you calculated.


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