As per requested: deriving Stokes' theorem from substituting this identity
Making substitution with
$ ∇ \times(\phi a) $
where $\phi$ is a scalar field, and $a$ is a vector field. Given that the identity is equal to:
$ ∇ \times(\phi a) = \phi (∇ \times a) + (∇\phi)\times a$
then derive stokes theorem:
$\int_{C}\phi dr = \iint_{S}n \times (∇\phi )dS$
for a surface $S$ with unit normal $n$ that is bounded by the closed curve $C$ , where the direction of $n$ obeys the usual convention.
What I've tried:
following from the hint
$$\nabla \times (\phi \mathbf{a}) \cdot \mathbf{n} = (\nabla \phi \times \mathbf{a})\cdot \mathbf{n} = (\mathbf{n} \times \nabla \phi) \cdot \mathbf{a}$$
substituting this into the theorem above gives,
$\int_{C} \phi \cdot \mathbf{a} dr=\iint_{S} (\mathbf{\hat{n}}\times\nabla\phi)\cdot\mathbf{a}dS$
Then taking out the constant a
$\mathbf{a}\cdot(\int_{C}\phi dr)=\mathbf{a}\cdot(\iint_{S}\mathbf{\hat{n}}\times(\nabla\phi)dS) $
since a is arbitrary we deduce that:
$\int_{C}\phi dr = \iint_{S}n \times (∇\phi )dS$
Best Answer
Hint:
If $\mathbf{a}$ is an arbitrary constant vector, then
$$\nabla \times (\phi \mathbf{a}) \cdot \mathbf{n} = (\nabla \phi \times \mathbf{a})\cdot \mathbf{n} = (\mathbf{n} \times \nabla \phi) \cdot \mathbf{a}$$