Prove $\int_{C}fdr=\int_{S}dS\times\nabla f$

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?

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

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

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

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