Relation between line integral of scalar function and surface integral

Question

I've seen the following identity on a book:

$$\int_{\partial S} f \, d\vec{\ell} = \iint_S d\vec{S} \times \nabla f$$

where $f$ is a scalar function and $\partial S$ is a closed curve.

I've been trying to prove it but I don't know where to start.

Answer 1

Let $\vec F$, the open surface $S$ and its boundary $\partial S$ meet the conditions of Stokes' Theorem. Then, we have

$$\oint_{\partial S} \vec F\cdot \hat t\,d\ell=\int_S \nabla\times\vec F\cdot \hat n\, dS$$

Now, let $\vec F=\hat x_i f$. Then, $\nabla \times \vec F=\nabla f\times \hat x_i$ and we have

$$ \oint_{\partial S}(\hat x_if) \cdot\,\hat t\,d\ell= \int_S (\nabla f\times \hat x_i)\cdot \hat n\, dS \tag1$$

Next, note that we have

$$\oint_{\partial S}(\hat x_if) \cdot\,\hat t\,d\ell=\hat x_i \cdot \oint_{\partial S}f\,d\vec\ell\tag2$$

and using the vector triple product we have

$$\begin{align} \int_S (\nabla f\times \hat x_i)\cdot \hat n\, dS=\hat x_i\cdot \int_S \hat n\times \nabla f\, dS\tag3 \end{align}$$

Substituting $(2)$ and $(3)$ into $(1)$ yields

$$\hat x_i \cdot \oint_{\partial S}f\,d\vec\ell=\hat x_i\cdot \int_S \hat n\times \nabla f\, dS\tag4$$

Since $(4)$ is true for all $i$, then we conclude that

$$\oint_{\partial S}f\,d\vec\ell=\int_S \hat n\times \nabla f\, dS$$

as was to be shown!