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Prove $\displaystyle \int_{C}fdr=\int_{S}dS\times\nabla f$. where $C=\partial S$ and the usual relationship between orientations hold.

Apply Stokes's theorem to $F=af$ where $a$ is an arbitrary constant vector.

From this identity and because $\nabla \times \mathbf{a = 0 }, $ thus $ \nabla\times F=0 + \nabla f \times a$.

Thus $(\nabla\times F)\cdot d\mathbf{ S }= (\nabla f \times a) \cdot d\mathbf{ S } = (d\mathbf{ S }\times \nabla f) \cdot a $, thanks to the answer below.

Then Stokes's theorem for arbitrary a implies $ \int_{C}f \mathbf{ a } \; d\mathbf{ r } = \iint_S (d\mathbf{ S }\times \nabla f) \cdot a $.

My concern: How do I proceed from here? Please explain steps in detail?

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Apply the Kelvin-Stokes theorem to the vector field $fa$:

$$\oint fa \cdot dr = \int (\nabla \times [af]) \cdot dS$$

Use vector calculus identities to write

$$\nabla \times [af] = f \nabla \times a + \nabla f \times a$$

Since $\nabla \times a = 0$, we get

$$\oint f a \cdot dr = \int (\nabla f \times a) \cdot dS$$

The triple product can be cyclically permuted, yielding

$$\oint fa \cdot dr = \int (dS \times \nabla f) \cdot a$$

Since $a$ is constant, it can be moved out of both integrals, the way you would move a constant scalar out of an integral:

$$a \cdot \oint f \, dr = a \cdot \int dS \times \nabla f$$

$a$ was chosen arbitrarily; this is true for all $a$ and thus we can "cancel" $a$. If one must think of it more rigorously, look at this expression above as a linear function of $a$. Take a gradient of that function with respect to $a$. The resulting gradients are equal on both sides. The result is

$$\oint f \, dr = \int dS \times \nabla f$$

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Remember that $$ (v\times w)\cdot z = \det(v,w,z) = \det (z,v,w) = (z\times v)\cdot w $$ and apply this to $v= \nabla f$, $w=a$, $z=ds$

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  • $\begingroup$ Thanks, but how to proceed after? $\endgroup$ – Greek - Area 51 Proposal May 30 '14 at 14:40
  • $\begingroup$ @LePressentiment what do you mean? true for all a..remove a $\endgroup$ – username Jun 3 '14 at 22:46

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