Use Stokes' theorem to show this integral relation.

Question

By applying Stokes’ theorem to the vector field $a×F$, where $a$ is an arbitrary constant vector and $F(r)$ is a vector field, show that

$$ \int_{C}dr\times F(r)=\int_{S}(dS\times \nabla)\times F $$

where the curve $C$ bounds the open surface $S$.

Stokes's theorem gives $\int_{S}(\nabla\times (a\times F))\cdot dS=\int_{C}(a\times F)\cdot dr$.

The right hand side is (being $a$ constant, and by permuting triple product) $a\cdot \int_{C}dr\times F(r)$, while the left hand side, being $a$ constant, is $\int_{S}(a(\nabla\cdot F)-(a\cdot \nabla)F)\cdot dS=a\cdot \int_{S}(\nabla\cdot F)\cdot dS-\int_{S}dS\cdot (a\cdot \nabla)F$.

I wonder if I can isolate the $a\cdot$ on this side like I did on the right hand side to obtain a relation without $a$. Or if I should use a particular value for $a$.

Answer 1

In index notation, $$ a\cdot \int_S(\nabla\cdot F)dS-\int_S dS\cdot (a\cdot \nabla)F =\int_S a_j dS_j\,(\partial_i F_i) - \int_S dS_j(a_i\partial_i)F_j\\ $$ Changing the dummy index in the second term, $$ \begin{align} &=\int_S a_j dS_j\,\partial_i F_i - \int_S dS_ia_j\partial_jF_i\\ &=a_j\int_S (dS_j\partial_iF_i-dS_i\partial_jF_i)\\ &=a\cdot\int_S (dS(\nabla\cdot F)-dS\cdot(\nabla F))\\ &=a\cdot\int_S (dS\times \nabla)\times F \end{align} $$