Show that for a vector field $\mathbf{A}$

$$\nabla \times(\boldsymbol{\nabla} \times \mathbf{A})=\boldsymbol{\nabla}(\boldsymbol{\nabla} \cdot \mathbf{A})-\nabla^{2} \mathbf{A}$$

Hence find an $\mathbf{A}(\mathbf{x})$, with $\boldsymbol{\nabla} \cdot \mathbf{A}=0$, such that $\mathbf{u}=\left(y^{2}, z^{2}, x^{2}\right)=\nabla \times \mathbf{A}$. [Hint: Note that $\mathbf{A}(\mathbf{x})$ is not defined uniquely. Choose your expression for $\mathbf{A}(\mathbf{x})$ to be as simple as possible.

Now consider the cone $x^{2}+y^{2} \leqslant z^{2} \tan ^{2} \alpha, 0 \leqslant z \leqslant h$. Let $S_{1}$ be the curved part of the surface of the cone $\left(x^{2}+y^{2}=z^{2} \tan ^{2} \alpha, 0 \leqslant z \leqslant h\right)$ and $S_{2}$ be the flat part of the surface of the cone $\left(x^{2}+y^{2} \leqslant h^{2} \tan ^{2} \alpha, z=h\right)$.

Using the variables $z$ and $\phi$ as used in cylindrical polars $(r, \phi, z)$ to describe points on $S_{1}$, give an expression for the surface element $d \mathbf{S}$ in terms of $d z$ and $d \phi$.

Evaluate $\int_{S_{1}} \mathbf{u} \cdot d \mathbf{S}$.

What does the divergence theorem predict about the two surface integrals $\int_{S_{1}} \mathbf{u} \cdot d \mathbf{S}$ and $\int_{S_{2}} \mathbf{u} \cdot d \mathbf{S}$ where in each case the vector $d \mathbf{S}$ is taken outwards from the cone?

What does Stokes theorem predict about the integrals $\int_{S_{1}} \mathbf{u} \cdot d \mathbf{S}$ and $\int_{S_{2}} \mathbf{u} \cdot d \mathbf{S}$ (defined as in the previous paragraph) and the line integral $\int_{C} \mathbf{A} \cdot d \mathbf{l}$ where $C$ is the circle $x^{2}+y^{2}=h^{2} \tan ^{2} \alpha, z=h$ and the integral is taken in the anticlockwise sense, looking from the positive $z$ direction?

Evaluate $\int_{S_{2}} \mathbf{u} \cdot d \mathbf{S}$ and $\int_{C} \mathbf{A} \cdot d \mathbf{l}$, making your method clear and verify that each of these predictions holds.