Vector field whose rotation is normal to a half ellipsoid

Question

I have a (special) half ellipsoid $$\tilde F=\left\{(x,y,z)\in\mathbb{R}\ \bigg|\ \frac{x^2}{a^2}+\frac{y^2}{a^2}+\frac{z^2}{b^2}=1,\ z\geq0\right\}$$ and need a vector field $f$ whose rotation is normal to that surface, so something like $$\begin{cases}\partial_yf_z-\partial_zf_y=\frac{x}{a^2}\\\partial_zf_x-\partial_xf_z=\frac{y}{a^2}\\\partial_xf_y-\partial_yf_x=\frac{z}{b^2}\end{cases}\qquad\forall (x,y,z)\in\tilde F $$ However, I'm not sure if something like this is even possible.

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In case you're wondering why I want this: I want to determine the surface area of the (full) ellipsoid $$F=\left\{(x,y,z)\in\mathbb{R}\ \bigg|\ \frac{x^2}{a^2}+\frac{y^2}{a^2}+\frac{z^2}{b^2}=1\right\}$$ Therefore I want to use the half ellipsoid, so $$v_2(F)=2v_2(\tilde F)$$ $\tilde F$ then is a compact subset of the submanifold $F$ with a smooth boundary (I'm not sure I translated this correctly from the german glatter Rand) $$\partial\tilde F=\left\{(x,y,0)\in\mathbb R^3\ \bigg|\ x^2+y^2=a^2\right\}$$ So one can apply the Kelvin–Stokes theorem $$\int_{\partial\tilde F}\langle f,d\vec s\rangle=\int_{\tilde F}\langle\mathrm{rot} f(x),\nu(x)\rangle dS(x)$$ Which looks similar to the surface I'm looking for, $$v_2(\tilde F)=\int_{\tilde F}dS(x)$$ So I merely have to achieve that $\langle\mathrm{rot} f(x),\nu(x)\rangle=1$ (and then calculate the line integral of the boundary). As $\nu$ is the normalized normal field to the ellipsoid, I can achieve this by requiring $\mathrm{rot} f(x)=\nu(x)$ - which leads to this question.

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It's kind of hard to explain (and draw) what I mean, but my idea was to modify $\nu$ outside of $\tilde F$ in a way, that the divergence is $0$, by letting the flow run in from the bottom - drawn in two dimensions: If $\nu$ then is normed (such that $\|\nu(x)\|=1\ \forall x$) the divergence should be $0$ (right?).

Answer 1

See this post here: for a fixed vector field $\mathbf{v}$ in a simply connected region $R$, we can find a vector field $\mathbf A$ such that $\mathbf v = \nabla \times \mathbf A$ if and only if $\nabla \cdot \mathbf v = 0$ in $R$.

Unfortunately, in this case the divergence $$\nabla \cdot \mathbf v = (1/a^2) + (1/a^2) + (1/b^2) $$ is never zero.