By applying the divergence theorem to the vector field $\bf a\times A$, where ${\bf a}$ is an arbitrary constant vector and ${\bf A}({\bf x})$ is a vector field, show that $\int_V {\bf \nabla \times \bf A} dV=−\int_S {\bf A \times} d{\bf S}$ , where the surface $S$ encloses the volume $V$.
So far I have done little more than just written down a few definitions and rearranged things a bit. There is a similar question to this that I would like to tackle and seeing a worked example of this I think would help with that.
Any help is appreciated
Thank you
By the divergence theorem,
$$\int_S \mathbf{a}\times\mathbf{A}\cdot \mathbf{n}\, d{S} = \int_V \nabla\cdot (\mathbf{a}\times \mathbf{A})\, dV$$
Note that the scalar triple product $\mathbf{a}\times \mathbf{A}\cdot \mathbf{n}$ is equal to $\mathbf{A}\times \mathbf{n}\cdot \mathbf{a}$. Thus
$$\int_S \mathbf{a}\times \mathbf{A}\cdot \mathbf{n}\, dS = \int_S \mathbf{A}\times \mathbf{n}\cdot \mathbf{a}\, dS = \int_S \mathbf{A}\times d\mathbf{S}\cdot \mathbf{a}$$
Now we express
$$\nabla \cdot (\mathbf{a}\times \mathbf{A}) = (\nabla \times \mathbf{a})\cdot \mathbf{A} - (\nabla \times \mathbf{A})\cdot \mathbf{a}$$
Then integrate over $V$ to obtain the expression
$$(\nabla \times \mathbf{a})\cdot \int_V \mathbf{A}\, dV - \int_V \nabla \times \mathbf{A}\, dV\cdot \mathbf{a}$$
Since $\mathbf{a}$ is constant, $\nabla \times \mathbf{a} = \mathbf{0}$. So the previous expression equals
$$-\int_V \nabla \times \mathbf{A}\, dV\cdot \mathbf{a}$$
Therefore
$$\left(\int_S \mathbf{A}\times d\mathbf{S}\right)\cdot \mathbf{a} = \left(-\int_V \nabla \times \mathbf{A}\, dV\right)\cdot \mathbf{a}$$
Since this equation holds true for every constant vector $\mathbf{a}$, the result follows (we can take $\mathbf{a} = \hat{\mathbf{i}}, \hat{\mathbf{j}}$, and $\hat{\mathbf{k}}$ to obtain equality of the $x$-, $y$-, and $z$-components of the vectors in parantheses).