How can one prove$\int_V\text{d}V\ \vec A=0$, given that $\vec A\cdot \vec{\textbf{n}}=0$ and $\nabla\cdot\vec A=0$ without using tensor analysis?

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

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

A standard solution is

  • $\int_V\nabla\cdot(\vec A\vec r )=\oint_S\text{d}\vec\sigma\cdot(\vec A\vec r )=\oint_S\text{d}\sigma\vec{\textbf{n}}\cdot(\vec A\vec r)=\oint_S\text{d}\sigma\ 0\times\vec r)=0$

However, does this solution mean that a man that haven`t learned tensor analysis can never solve the problem? I wonder whether there are any other solutions without tesors.

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