I am a little stuck on the following problem:

Use Stokes's Theorem to show that

$$\oint_{C} y ~dx + z ~dy + x ~dz = \sqrt{3} \pi a^2,$$

where $C$ is the suitably oriented intersection of the surfaces $x^2 + y^2 + z^2 = a^2$ and $x + y + z = 0$.

OK, so Stokes's Theorem tells me that:

$$\oint_{C}\vec{F} \cdot d\vec{r} = \iint_{S}\operatorname{curl} \vec{F} \cdot \vec{N} ~dS$$

I have calculated:

$$\operatorname{curl} \vec{F} = -\vec{i} - \vec{j} - \vec{k}.$$

I then figured that on the surface $S$ we must have:

$$\vec{N}dS = \vec{i} + \vec{j} + \vec{k}dxdy$$

since this follows from the equation of the given plane.

However this will then give me:

$$\operatorname{curl} \vec{F} \cdot \vec{N} = -1 -1 -1 = -3$$

And thus I would get, if project this onto the $xy$-plane:

$$\iint_{S} \operatorname{curl} \vec{F} \cdot \vec{N} ~dS = -3 \iint_{A} dA = -3 \pi a^2$$

which is obviously not correct.

I would greatly appreciate it if someone could help me with this. I actually had multivariable calculus a few years ago, and I know that I knew this stuff then. However, now that I need it again I notice that I've become quite rusty.

Thanks in advance :)

  • 5
    $\begingroup$ You should normalize $N$! $\endgroup$
    – Fabian
    Jul 17, 2012 at 19:07
  • $\begingroup$ Thanks. But I am slightly confused, as my book writes that $\vec{N}dS = \pm \frac{\nabla G(x,y,z)}{G_{3}(x,y,z)}$ $dxdy$. (Although I see from this that I should of course drop the minus-sign in front of my given answer, since we here are using the upward normal). Yet this would then give me the answer $3 \pi a^2$ $\endgroup$
    – Kristian
    Jul 17, 2012 at 19:18
  • $\begingroup$ @Kristian: If you normalize $N$, you get $\frac{1}{\sqrt{3}}(\vec{i}+\vec{j}+\vec{k})$. After pulling out the normalizing factor (and correctly orienting the curve) you will get $\sqrt{3}\pi a^2$, just as expected. $\endgroup$ Jul 17, 2012 at 19:29
  • $\begingroup$ Thank, Arturo. I see of course that when I normalize $\vec{N}$ I get what you write. But isn't $dS = \sqrt{3}$ so that $\vec{N}dS = \frac{1}{\sqrt{3}}(\vec{i} + \vec{j} + \vec{k}) \cdot \sqrt{3} = \vec{i} + \vec{j} + \vec{k}$? $\endgroup$
    – Kristian
    Jul 17, 2012 at 19:35
  • $\begingroup$ I thought like that ,too... $\endgroup$ Jul 17, 2012 at 19:36

1 Answer 1


This is an effort to get this question off from the unanswered queue. Also here I use a rather general approach than the classical Kelvin-Stokes theorem.

Stokes theorem reads: $$ \int_{\partial M} \omega = \int_{M} d\omega. $$ Hence $$ \int_C y\,dx + z\,dy +x\,dz = \pm \int_S d(y\,dx + z\,dy +x\,dz) \\ = \pm\int_S dy\wedge dx + dz\wedge dy + dx\wedge dz. $$ For simplicity we choose $S$ to be planar surface bounded by the sphere, not the hemisphere bounded by the plane. On the plane $z = -x-y$, hence above integral is $$ \pm 3\int_S dx\wedge dy.\tag{$\star$} $$ Using the natural parametrization of the plane, this integral can be interpreted as the area of projected area $S$ on to the $xy$-plane (or notice $\star dz = dx\wedge dy$). The intersection of $x+y+z = 0$ with $x^2 + y^2 + z^2 =a^2$, projected onto the $xy$-plane, is an ellipse.

The end points for this projected ellipse are: minor axis end points are projection of $(-1/\sqrt{6},-1/\sqrt{6},2/\sqrt{6} )a$, and $(1/\sqrt{6},1/\sqrt{6},-2/\sqrt{6})a$, achieved when we set $x=y$. We can compute the length of the minor axis is $b = \sqrt{3}a/3$. The major axis end points are achieved by setting $x+y=0$, $(\sqrt{2}/2, -\sqrt{2}/2, 0)a$ and $(\sqrt{2}/2, -\sqrt{2}/2, 0)a$, the length of the major axis is just $a$. Hence the project area is $\sqrt{3}\pi a^2/3$ and plugging back to $(\star)$ yields $$ \int_C y\,dx + z\,dy +x\,dz = \pm \sqrt{3} \pi a^2. $$ The sign depends on $C$ is chosen to be rotated counter-clockwisely or clockwisely with respect to the normal vector to the plane $x+y+z=0$.

To address your own question:

Why I got $\displaystyle \iint_{S} \operatorname{curl} \vec{F} \cdot \vec{n} ~dS = -3 \iint_{A} dA = -3 \pi a^2$ if I use $dS = \sqrt{3}dA$ so that $n\,dS = (1/\sqrt{3},1/\sqrt{3},1/\sqrt{3})\sqrt{dS} = (1,1,1)$?

This is not correct, for that if you use the area element $dS = \sqrt{3} \,dx dy = \sqrt{3} \,dA$, the new $A$ is the projected ellipse on the $xy$-plane, having area $\sqrt{3}\pi a^2/3$ (please see the argument above), not having the same area $\pi a^2$ with $S$.

Standard way using Kelvin-Stokes as Arturo Magidin pointed out in the comments: choosing $C$ rotates counter-clockwisely with respect to the unit vector $n = -(1,1,1)/\sqrt{3}$ normal to the plane $x+y+z=0$: $$ \int_C y\,dx + z\,dy +x\,dz = \int_{S} \nabla \times (y,z,x)\cdot n\,dS \\ = \int_{S} (-1,-1,-1)\cdot (-1,-1,-1)/\sqrt{3} \,dS = \sqrt{3}|S| = \sqrt{3}\pi a^2. $$


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