Stoke's Theorem Theory Question

Question

If $C$ is the boundary of a surface $S$ and $\phi$ and $\psi$ are arbitrary smooth scalar fields, Show that $$\int_C \phi \nabla \psi \cdot dr = - \int_C \psi \nabla \phi \cdot d r = \iint_S (\nabla \phi \times \nabla \psi) \cdot n dS$$

I've tried working backwards, letting $\phi$ and $\psi$ be arbitrary and then generating something, but I just get a mess of symbols. Can't seem to figure it out.

Answer 1

Use $$\text{curl} (fX) = \nabla \times fX = f\, \nabla \times X + \nabla f \times X$$ and $$\text{curl} (\nabla f) = \nabla \times \nabla f = 0 $$

for every smooth scalar field (function) $f$ and vector field $X$.

Then Stokes Theorem implies \begin{align*} \int_C \phi \nabla \psi \cdot dr & = \iint_S (\nabla\times (\phi \nabla \psi)) \cdot n dS = \iint_S (\nabla \phi \times \nabla\psi + \phi \nabla \times \nabla\psi) \cdot n dS = \\ % & = \iint_S (\nabla \phi \times \nabla\psi) \cdot n dS = -\iint_S (\nabla \psi \times \nabla\phi) \cdot n dS = - \int_C \psi \nabla \phi \cdot dr \end{align*}