Differential surface element and nabla operator

Question

If we have the vector field $\vec{u}=\vec{A}\times \vec{v}$, where $\vec{A}=\text{const.}$ and we integrate over some closed curve, by using Stokes' theorem we get: $$ \begin{align} \oint_{\partial S} \vec{u} \cdot d\vec{r} &= \iint_S \nabla\times(\vec{A}\times\vec{v}) \cdot d\vec{S} \\ &= \iint_S \bigl( \vec{A}(\nabla\cdot\vec{v}) - \vec{A} \cdot (\nabla \otimes\vec{v}) \bigr) \cdot d\vec{S} \\ &= \vec{A} \cdot \iint_S(\nabla\cdot\vec{v}) \cdot d\vec{S} - (\nabla \otimes\vec{v})\cdot d\vec{S} \end{align} $$ And at this point my textbook says that: $$ (\nabla\otimes\vec{v}) \cdot d\vec{S} = \nabla(\vec{v} \cdot d\vec{S}) $$ I just can't understand why this equality holds when the differential vector $d\vec{S}$ depends on the coordinates and hence in not a constant generally. Can you give me a proof of why is that the case?

Answer 1

We know that $$\vec u\otimes\vec w=\begin{bmatrix}u_1\\u_2\\u_3\end{bmatrix}\begin{bmatrix}w_1&w_2&w_3\end{bmatrix};\\ \vec u\cdot\vec w=\begin{bmatrix}u_1&u_2&u_3\end{bmatrix}\begin{bmatrix}w_1\\w_2\\w_3\end{bmatrix};\\ \nabla\phi=\begin{bmatrix}\displaystyle\frac{\partial}{\partial x_1}\\\displaystyle\frac{\partial}{\partial x_2}\\\displaystyle\frac{\partial}{\partial x_3}\end{bmatrix}\begin{bmatrix}\phi\end{bmatrix}.$$ Thus, we have \begin{align} (\nabla\otimes\vec v)\cdot d\vec S&=\left(\begin{bmatrix}\displaystyle\frac{\partial}{\partial x_1}\\\displaystyle\frac{\partial}{\partial x_2}\\\displaystyle\frac{\partial}{\partial x_3}\end{bmatrix}\begin{bmatrix}v_1&v_2&v_3\end{bmatrix}\right)\begin{bmatrix}dS_1\\dS_2\\dS_3\end{bmatrix}\\ &=\begin{bmatrix}\displaystyle\frac{\partial}{\partial x_1}\\\displaystyle\frac{\partial}{\partial x_2}\\\displaystyle\frac{\partial}{\partial x_3}\end{bmatrix}\left(\begin{bmatrix}v_1&v_2&v_3\end{bmatrix}\begin{bmatrix}dS_1\\dS_2\\dS_3\end{bmatrix}\right)\\ &=\begin{bmatrix}\displaystyle\frac{\partial}{\partial x_1}\\\displaystyle\frac{\partial}{\partial x_2}\\\displaystyle\frac{\partial}{\partial x_3}\end{bmatrix}\begin{bmatrix}\vec v\cdot d\vec S\end{bmatrix}\\ &=\nabla(\vec v\cdot d\vec S). \end{align} Therefore, $$(\nabla\otimes\vec v)\cdot d\vec S=\nabla(\vec v\cdot d\vec S).$$