electromagnetism - How can one prove $int_Vtext{d}V vec A=0$, given that $vec Acdot vec{textbf{n}}=0$ and $nablacdotvec A=0$, without using tensor analysis?

In the course of learning electrodynamics, I was asked to solve the problem following:

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  $\vec A$ is a vector which satisfies $\vec A\cdot \vec{\textbf{n}}=0$, where $\vec{\textbf{n}}$ is the normal vector of the surface of the volume $V$. Besides, $\nabla\cdot\vec A=0$ within the volume $V$. Please prove $$\int_V\text{d}V\ \vec A=0.$$

  
  


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  A standard solution is $$\int_V\text{d}V \vec A=\int_V\text{d}V\,\nabla\cdot(\vec A\vec r )=\oint_S\text{d}\vec\sigma\cdot(\vec A\vec r )=\oint_S\text{d}\sigma\,\vec{\textbf{n}}\cdot(\vec A\vec r)=\oint_S\text{d}\sigma\ (0\times\vec r)=0.$$

  
  

However, does this solution mean that a person that has never learned tensor analysis can never solve the problem? I wonder whether there are any other solutions without using tensors.

Thank you for your reading the question. Waiting for your excellent answers.

Answer

"Tensor" analysis is just fancy language, and you can do this just fine without it. The essence of tensor analysis (or at least one way of looking at said essence; see this answer for more on that) is to do things component by component. Since $$ \int_V\text{d}V\ \mathbf A=\sum_j \hat{\mathbf e}_j\int_V\text{d}V\ (\hat{\mathbf e}_j\cdot\mathbf A), $$ it is sufficient to just consider the integral of the scalar quantity $A_j=\hat{\mathbf e}_j\cdot\mathbf A$. In this language, the proof is a reformulation of what you've given: because $$ \nabla \cdot(x_j\mathbf A) = (\nabla x_j)\cdot\mathbf A + x_j \nabla\cdot\mathbf A = \hat{\mathbf e}_j\cdot\mathbf A, $$ we can write \begin{align} \int_V\text{d}V\ (\hat{\mathbf e}_j\cdot\mathbf A) & = \int_V\text{d}V\ \nabla \cdot(x_j\mathbf A) \\ & = \oint_S\text{d}\mathbf S\cdot(x_j\mathbf A), \end{align} which vanishes because $\mathbf A$ and $\mathrm d\mathbf S$ are orthogonal, and you're done. See? Easy! The tensor-analysis layer is just some fancy language on top to make everything come together slightly more coherently, but it is fundamentally the same proof.