# Stokes Theorem (Application)

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 $C_1$ & $C_2$. $$\int_{C_1} \vec A.dr_1 -\int_{C_2} \vec A.dr_2$$

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?

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