0
$\begingroup$

As far as i can understand, Stokes Theorem says that it doesn't matter which surface $S$ with boundary $C$ (closed curve) i take, the circulation of a vector field $F$ through that surface is $\int_C F \cdot dr$

But, let say i have the curve $x^2+y^2=1$ and $F(x,y,z)=(y,x,z)$ with $curl\ F=0$ so $\int_C F \cdot dr=\iint_S \ curl\ F\cdot dS=0$. This is correct for the circle $x^2+y^2\le1$ but for the paraboloid $z=1-x^2-y^2,\ \ z \ge 0$ the surface integral is $\pi/2$ and not $0$ so Stokes Theorem is not correct?

Is there something that I don't understand correctly or maybe i'm missing a condition for the theorem to work? I appreciate your time to help me.

$\endgroup$
  • 1
    $\begingroup$ How do you get $\pi/2$ for the surface integral? $\text{curl}(F)=0$ so its surface integral through any surface must be zero. $\endgroup$ – kccu Dec 6 '17 at 22:05
  • $\begingroup$ Also your first sentence is not quite correct. Stokes' Theorem says that the surface integral of $\text{curl}(F)$ (not $F$) through $S$ is equal to $\int_C F \cdot dr$. Perhaps this is the source of your confusion. $\endgroup$ – kccu Dec 6 '17 at 22:07
  • $\begingroup$ Oh!... that's true. Everything makes much more sense now, thank you! $\endgroup$ – BadEnglishSorry Dec 6 '17 at 22:34

Your Answer

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

Browse other questions tagged or ask your own question.