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 '17 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 '17 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 '17 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 '17 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 '17 at 11:14

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 '17 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: $z=r\cos \phi, x=r\sin \phi \cos\theta, y=r\sin \phi\sin \theta$ $\endgroup$ Aug 18 '17 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 '17 at 17:35

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.