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Stock`s theorem $$\oint\limits_C {{\bf{a}} \cdot {\bf{dr}}} = \iint\limits_S {\nabla \times {\bf{a}}\, \cdot {\bf{n}}dA}$$

Substituting ${\bf{a}} = {\bf{f}} \times {\bf{c}}$ we find that $$\oint\limits_C {{\bf{dr}} \times {\bf{f}}} = \iint\limits_S {\left( {{\bf{n}} \times \nabla \,} \right) \times {\bf{f}}dA}$$ since ${\bf{n}} \cdot \left( {\nabla \times \left( {{\bf{f}} \times {\bf{c}}} \right)} \right) = {\bf{c}} \cdot \left( {\left( {{\bf{n}} \times\nabla } \right) \times {\bf{f}}} \right) % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYfdmGievaebbnrfi % fHhDYfgasaacH8srps0lbbf9q8WrFfeuY-Hhbbf9v8qqaqFr0xc9pk % 0xbba9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9 % Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWHUbGaey % yXIC9aaeWaaeaadaWhcaqaaiabgEGirdGaay51GaGaey41aq7aaeWa % aeaacaWHMbGaey41aqRaaC4yaaGaayjkaiaawMcaaaGaayjkaiaawM % caaiabg2da9iaahogacqGHflY1daqadaqaamaabmaabaGaaCOBaiab % gEna0oaaFiaabaGaey4bIenacaGLxdcaaiaawIcacaGLPaaacqGHxd % aTcaWHMbaacaGLOaGaayzkaaaaaa!55F5! $

I can`t prove last equation, please help,

c is constant vector, n is unit vector, a and f are vector functions

Igor Fomenko
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  • $f$ is a function? $n$ a unit vector? What is $n\times\nabla$? In any case it appears be a trivial (but long) calculation. – Martín-Blas Pérez Pinilla Dec 17 '18 at 17:47
  • Yes, f is vector function, n is unit vector, $${\bf{n}} \times \nabla % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOBaiabgE % na0kabgEGirdaa!3A8A! $$ is a cross product. – Igor Fomenko Dec 18 '18 at 20:41
  • And what is the "cross product" of vector $\times$ operator? – Martín-Blas Pérez Pinilla Dec 19 '18 at 15:03
  • It`s a vector operator ${\bf{n}} \times \nabla = {n_i}{{{\bf{\hat e}}}_i} \times {{{\bf{\hat e}}}_j}{\partial _j} = {{{\bf{\hat e}}}_k}{\varepsilon _{ijk}}{n_i}{\partial _j}$ – Igor Fomenko Dec 20 '18 at 08:09

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