Given vector field $F=\langle xyz,x,e^{xy}\cos z\rangle$ and a semi-sphere $x^2+y^2+z^2\le 1$ over $z=0$ and a normal vector $n$ to the surface, calculate $\iint \operatorname{curl}F\cdot ndS$.

Because $$\operatorname{curl}F=\langle xe^{xy}\cos z,xy−ye^{xy}\cos z,1-xz\rangle$$ it looks like a nightmare to calculate $\iint \operatorname{curl}F\cdot ndS$ directly.

So I thought we can define a new field $G=\operatorname{curl}F$. Also let $S$ be the surface of semi-sphere and $B$ be the surface which closes the semi-sphere from the bottom.

Then $$\iint_S \operatorname{curl}F\cdot ndS=\iint_{S\cup B} \operatorname{curl}F\cdot ndS-\iint_B \operatorname{curl}F\cdot ndS\stackrel{\text{by Gauss theorem}}{=}\\ =\iiint_{S\cup B}\operatorname{div}(\operatorname{curl}F)-\iint_B \operatorname{curl}F\cdot ndS$$ Because $\operatorname{div}(\operatorname{curl}F)=0$ always then we just need to calculate the integral over $B$ and it's much easier because the normal unit vector is essentially the $z$ axis in the negative direction. $$ \iint_{B} \operatorname{curl}F\cdot ndS=\iint_{ B} \operatorname{curl}F\cdot \langle0,0,-1\rangle dS=\iint_{B} (xz-1) dS\\ \stackrel{\text{z=0}}{=}\iint_{B}-1 dS\stackrel{\text{circle area}}{=}-\pi $$ Thus: $$ \iint_S \operatorname{curl}F\cdot ndS=\pi. $$ Is this in the right direction? I'm particularly not sure if $\operatorname{div}(\operatorname{curl}F)=0$ in this case.

  • $\begingroup$ There are a few mistakes in here. First of all your calculation for $curl(F)$ is wrong. see this link (you need to copy paste it): wolframalpha.com/input/?i=curl(xyz,x,x%5E(x*y)*cos(x)) $\endgroup$
    – Maik Pickl
    Aug 14, 2017 at 11:06
  • $\begingroup$ Then you are supposed to use the Stokes-Kelvin Theorem for curl: en.wikipedia.org/wiki/… $\endgroup$
    – Maik Pickl
    Aug 14, 2017 at 11:08
  • 1
    $\begingroup$ If you knew F, you could apply stokes theorem to get a path integral around the edge of the half-sphere. This immediately simplifies things because you would have z=0 around the path. Alas, I studied this fifty years ago, and I have no idea how to get F(x,y,z) from the curl. But if you could, then Pickl 1 comment is the way to go. $\endgroup$ Aug 14, 2017 at 11:08
  • $\begingroup$ The boundary of the semisphere will be a circle in the x-y-plane. And you don't need to calculate $curl(F)$ for this at all. Try it and let me know if you need further help. $\endgroup$
    – Maik Pickl
    Aug 14, 2017 at 11:09
  • $\begingroup$ @richard1941 We had the same idea. :) Just want to point out that $F$ is given in the first line of the question. :) $\endgroup$
    – Maik Pickl
    Aug 14, 2017 at 11:14

1 Answer 1


Assume ${\bf F}$ as given, and assume that the upper hemisphere $S: \>x^2+y^2+z^2=1,\ z\geq0$ is oriented upwards. The surface $S$ has a boundary cycle $\partial S$ which is the unit circle in the $(x,y)$-plane, oriented counterclockwise.

We are told to compute the flux integral $$\Phi:=\int_S{\rm curl}({\bf F})\cdot{\bf n}\>d\omega\ .\tag{1}$$ This computation can be performed in three ways:

(i) Compute ${\bf C}:={\rm curl}({\bf F})$ as a function of $x$, $y$, $z$, use the parametric representation $${\bf r}(\phi,\theta):=\bigl(\cos\theta\cos\phi,\cos\theta\sin\phi,\sin\theta\bigr)$$ ($\phi$ and $\theta$ are GPS coordinates) for $S$ and compute the surface integral as given: $$\Phi=\int_0^{\pi/2}\int_0^{2\pi}{\bf C}\bigl({\bf r}(\phi,\theta)\bigr)\cdot\bigl({\bf r}_\phi\times{\bf r}_\theta)\>d\phi\>d\theta\ .$$ (ii) Use Stokes' theorem to convert $(1)$ into a line integral along $\partial S$; then compute this line integral: $$\Phi=\int_{\partial S}{\bf F}\cdot d{\bf r}\ .\tag{2}$$ Going this way you don't even have to compute ${\bf C}$, but you need to parametrize $\partial S$: $${\bf r}(\phi)=(\cos\phi,\sin\phi,0)$$ and plug this into $(2)$.

(iii) You have chosen a third way, namely using Gauss' theorem. This theorem deals with a three-dimensional solid $B$ and its boundary surface $\partial B$. We define $B$ to be the half ball bounded by $S$ and the unit disc $U$ in the $(x,y)$-plane oriented downwards. Gauss' theorem then gives $$\int_{\partial B}{\bf C}\cdot{\bf n}\>d\omega=\int_B{\rm div}({\bf C})\>{\rm dvol}=0\ ,$$ since ${\rm div}\circ{\rm curl}=0$. From $\partial B=S+U$ it follows that $$\Phi=-\int_U {\bf C}\cdot{\bf n}\>{\rm d}(x,y)=\int_U C_3(x,y,0){\rm d}(x,y)\ .$$ It remains to correctly compute $C_3$, which I leave to you.

  • $\begingroup$ To add a variant combining the second and third way, you could also use Stokes twice and note that $\partial S = - \partial U$, in order to arrive at the same integral over $U$. $\endgroup$
    – mlk
    Aug 18, 2017 at 9:21
  • $\begingroup$ I have a question on the first method you described, specifically the parametrization ${\bf r}(\phi,\theta):=\bigl(\cos\theta\cos\phi,\cos\theta\sin\phi,\sin\theta\bigr)$. What kind of coordinates are those? Seems similar to spherical ones but they are defined differently: $z=r\cos \phi, x=r\sin \phi \cos\theta, y=r\sin \phi\sin \theta$ $\endgroup$
    – Yos
    Aug 18, 2017 at 16:47
  • 1
    $\begingroup$ @user123429842: Unfortunately there is no standard convention about spherical coordinates. I use geographical coordinates: the latitude $\theta$ and the longitude $\phi$, as the GPS system does. Having $\theta=0$ at the equator reproduces the symmetry of the sphere in a nice way. If you are studying the motion of a spinning top you might prefer $\theta=0$ at the north pole. $\endgroup$ Aug 18, 2017 at 17:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .