Show that $\oint_{\partial S} \psi \,\mathrm{d}\ell = -\iint_S \nabla \psi \times \mathrm{d}S$.

Question

I want to prove following identity:

$$\oint_{\partial S} \psi \,\mathrm{d}\ell = -\iint_S \nabla \psi \times \mathrm{d}S. $$

Here $\psi$ is a scalar function. I'm in trouble about understanding this expression. I haven't seen both sides of equality anywhere. The divergence theorem and the Stokes' theorem doesn't seem related to the question.

Answer 1

You can use the fact:

Two vectors $u, v$ are equal iff $u\cdot k = v\cdot k$ for all vector $k$.

For any constant vector $k$, we have

$$\require{cancel}\begin{align}k \cdot \oint_{\partial S} \psi d\ell &= \oint_{\partial S} (\psi k) \cdot d\ell\\ \color{blue}{\text{Stoke's theorem }\rightarrow}&= \int_{S} (\nabla \times (\psi k)) \cdot dS\\ &= \int_{S} ((\nabla \psi) \times k + \psi \color{red}{\cancelto{0}{\color{gray}{(\nabla \times k )}}})\cdot S\\ &= -\int_{S} k \cdot (\nabla \psi \times dS )\\ &= -k\cdot \int_S \nabla \psi \times dS \end{align} $$ Since this is true for all $k$, you get

$$\oint_{\partial S} \psi d\ell = -\int_S \nabla \psi \times dS$$

Answer 2

The given vector identity follows from Stokes' Theorem: \begin{align}\oint_{\partial S} \psi\, d{\boldsymbol\ell}&= \mathbf{i}\oint_{\partial S} (\psi,0,0)\cdot d{\boldsymbol\ell}+ \mathbf{j}\oint_{\partial S} (0,\psi,0)\cdot d{\boldsymbol\ell}+ \mathbf{k}\oint_{\partial S} (0,0,\psi)\cdot d{\boldsymbol\ell}\\ &\stackrel{\text{Stokes}}{=}\mathbf{i}\iint_S(0,\psi_z,-\psi_y)\cdot d{\boldsymbol S}+\mathbf{j}\iint_S(-\psi_z,0,\psi_x)\cdot d{\boldsymbol S} +\mathbf{k}\iint_S(\psi_y,-\psi_x,0)\cdot d{\boldsymbol S}\\ &=\mathbf{i}\iint_S(\psi_z (d{\boldsymbol S})_y-\psi_y(d{\boldsymbol S})_z) +\mathbf{j}\iint_S(-\psi_z(d{\boldsymbol S})_x+\psi_x (d{\boldsymbol S})_z)\\ &\qquad\qquad+\mathbf{k}\iint_S(\psi_y (d{\boldsymbol S})_x-\psi_x(d{\boldsymbol S})_y)\\ &=\iint_S d{\boldsymbol S}\times (\psi_x,\psi_y,\psi_z) =-\iint_S \nabla \psi\times d{\boldsymbol S}.\end{align}