Gradient and differential surface element

Question

In my textbook There is the following step made when dealing with a certaion surface integral over a random surface: $$\int_S(\nabla\otimes\vec{v})\cdot d\vec{S}=\int_S\nabla(\vec{v}\cdot d\vec{S})$$ And after that it says - "We used the fact that $\nabla$ doesnt act on $d\vec{S}$, when there is a contraction(dot product)."$$$$ First of all the differentail vector field $d\vec{S}$ is defined on a subset of $\mathbb{R}^3$ (on the surface of integration) so the gradient operator shouldn't make any sense since if we make a small step in almost any direction we go off the surface and so the vector $d\vec{S}$ is going to be undefined in that point. For it to work how they said, the gradient of that field should be zero. WHY? The second question is why does this work exactly when $d\vec{S}$ is in contraction? Why wouldn't it work for example in a cross product? maybe that would explain my first question? I have no idea.

Answer 1

On p. 107 in line 2460 I do not see your formula but instead Stokes' theorem for $\vec{u}=\vec{A}\times\vec{v}$ where $\vec{A}$ must be constant followed by a well-known expression for $\nabla\times(\vec{A}\times\vec{v})\,:$ \begin{align*} \oint_{\partial S}\vec{u}\cdot d\vec{r}&=\int_S\nabla\times(\vec{A}\times\vec{v})\cdot d\vec{S}=\int_S\Big(\vec{A}(\nabla\cdot\vec{v})-(\vec{A}\cdot\nabla)\vec{v}\Big)\cdot d\vec{S}\,. \end{align*} The author then says that the last integral can be written as \begin{align*} \vec{A}\cdot \int_S(\nabla\cdot\vec{v})\,d\vec{S}-\vec{A}\cdot\int_S\nabla(\vec{v}\cdot d\vec{S})\,. \end{align*} I think only the identity $$ \int_S(\vec{A}\cdot\nabla)\vec{v}\cdot d\vec{S}=\vec{A}\cdot\int_S\nabla(\vec{v}\cdot d\vec{S}) $$ needs a bit of thinking. When the LHS is expressed using a differential form we get \begin{align} &\int_S \Big(A_x\partial_x v_x+A_y\partial_yv_x+A_z\partial_zv_x\Big)\,dy\wedge dz\\ &+\int_S \Big(A_x\partial_x v_y+A_y\partial_yv_y+A_z\partial_zv_y\Big)\,dz\wedge dx\\ &+\int_S \Big(A_x\partial_x v_z+A_y\partial_yv_z+A_z\partial_zv_z\Big)\,dx\wedge dy\\ &=A_x\int_S\partial_xv_x\,dy\wedge dz+\partial_xv_y\,dz\wedge dx+\partial_xv_z\,dx\wedge dy\\ &+A_y\int_S\partial_yv_x\,dy\wedge dz+\partial_yv_y\,dz\wedge dx+\partial_yv_z\,dx\wedge dy\\ &+A_z\int_S\partial_zv_x\,dy\wedge dz+\partial_zv_y\,dz\wedge dx+\partial_zv_z\,dx\wedge dy\,.\tag{1} \end{align} Keeping in mind that $\vec{v}\cdot d\vec{S}$ is in differential form notation $$ \omega=v_x\,dy\wedge dz+v_y\,dz\wedge dx+v_z\,dx\wedge dy $$ I think there is nothing wrong with writing the long expression (1) as $\vec{A}\cdot\int_S\nabla(\vec{v}\cdot d\vec{S})\,.$ A more modern notation could be obtained using the Lie derivative of $\omega\,:$ \begin{align} {\cal L}_A(\omega)&=d(i_A(\omega))+i_A(d\omega)\, \end{align} which is a $2$-form representing the derivative of $\omega$ along the constant vector field $$ A=A_x\partial_x+A_y\partial_y+A_z\partial_z=\vec{A}\cdot\nabla\,. $$ I found it a good exercise to show that (1) can also be written neatly as $$ \int_S{\cal L}_A(\omega)\,. $$