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Statement: If a vector field R is irrotational then a line integral is independent of path.

Proof. Let $\nabla$ $\times$ $\vec A=0$ in $R$ consider the difference of two line integral from the point $r_0$ to $r$ along two curves $C_1$ & $C_2$.

enter image description here $$\int_{C_1} \vec A.dr_1 -\int_{C_2} \vec A.dr_2$$

where $r_1$ is integration variable to distinguish it from the limit of integration $r$ &$r_0$

now $$\int_{C} \vec A.dr_1 = \int_{C_1} \vec A.dr_1 -\int_{C_2} \vec A.dr_2$$

from Stokes theorem $\int_{C} \vec A.dr_1$ =$\int_{S} \nabla \times \vec A.\vec ds $

$\nabla \times \vec A=0 $

so $\int_{C} \vec A.dr_1=0$

Is there any mistake?

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  • 1
    $\begingroup$ Welcome to Math.SE. Please typeset your question. These photos are barely readable. Here is a tutorial. $\endgroup$ – Ayman Hourieh Feb 2 '14 at 14:14
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    $\begingroup$ One thing that is wrong is that the man's name was Stokes, not Stoke. $\endgroup$ – Hans Lundmark Feb 2 '14 at 14:18
  • $\begingroup$ It depends on what level student you are. How do you know an arbitrary closed curve bounds an orientable surface $S$? $\endgroup$ – Ted Shifrin Feb 2 '14 at 14:23
  • $\begingroup$ OK sorry. Any mathematical mistake? $\endgroup$ – Jameel Feb 2 '14 at 14:23
  • $\begingroup$ surface integral of curl A over an open surface S is bounded by C $\endgroup$ – Jameel Feb 2 '14 at 14:40

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