I have the following problem and I'm not able to solve it. Given $F(x,y,z) = (1-2z, 0, 2y)$, calculate the line integral of C, where C is the contour of the surface. $S=\{(x,y,z) / x\geq 0, y \geq 0, z\geq 0, z=1-x^2 - y^2\}$

Using Stokes Theorem:

$$\int _S \nabla x F\cdot dS = \int_C F\cdot dS $$

I have calculated the Curl $\nabla x F = (2,-2,0)$
I also parametrized the surface with $T(u,v)=(u,v,1-u^2-v^2)$
So my normal vector is $T_u x T_v=(2u,2v,1)$
$$\int_CF\cdot dS=\int _S \nabla x F\cdot dS =\int_0^1\int_0^1 (4u-4v)dudv=0$$

But C is a closed curve!! So if the line integral over a closed curve is cero, by the conservative vector field theorem, the curl of the field should ALSO be cero!! WTF, is going on? PLEASE HELP.

  • $\begingroup$ What makes you think $F$ is a conservative vector field? $\endgroup$ – Boots Mar 9 at 1:04
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    $\begingroup$ Does the integral vanish over every closed curve, or just this particular one? $\endgroup$ – amd Mar 9 at 1:06
  • $\begingroup$ Boots. The conservative vector field says all of the following are equivalent: i) The line integral over a closed curve is zero ii) The line integral from A to B is the same, no matter the curve you take iii) $\nabla x F =0$ iv) it exist a potential function f such that $\nabla f = F$ $\endgroup$ – Clock Maker Mar 9 at 1:58
  • $\begingroup$ amd - It vanish over this particular curve that I know. I haven't checked for EVERY particular curve. $\endgroup$ – Clock Maker Mar 9 at 2:03
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    $\begingroup$ I got it, it does vanish over this particular curve, that doens't mean it vanish for EVERY curve. What a moron, thanks amd, I freaking love you. $\endgroup$ – Clock Maker Mar 9 at 2:09

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