Counter clockwise flux integral There's a function $\ \varphi :(0,+\infty) \to \Bbb R$ and another function $\ u:\Bbb R^{2} \to \Bbb R$ defined as $\ u(x,y):=\varphi(x^2+4y^2)$. For some $\ t>0$, let $\ E_{t} = \{{(x,y): x^2+4y^2=t}\}. $
Now trying to find the counter clockwise flux $\ {\unicode{x2233}}_{E_{t}} \nabla u \cdot \bar n ds$ of $\ \nabla u$ across $\ E_{t}$ and find all functions $\ \varphi$ so that this flux is independent of t. 
Any ideas?
 A: Consider the function
$$v(x,y):=x^2+4y^2\ .$$ The set $E_t:=\{(x,y)\>|\>v(x,y)=t\}$, $t>0$, is an ellipse bounding an elliptical domain $D_t$ of area ${\pi\over2} t$. Assume that ${\bf n}$ is the outward normal of $E_t$. Then by Gauss' divergence theorem we have
$$\int_{E_t}\nabla v\cdot{\bf n}\>{\rm d}s=\int_{D_t}{\rm div}(\nabla v)\>{\rm d}(x,y)\ .$$
Since ${\rm div}(\nabla v)=\Delta v=v_{xx}+v_{yy}\equiv10$
we obtain
$$\int_{E_t}\nabla v\cdot{\bf n}\>{\rm d}s=10\>{\rm area}(D_t)=5\pi t\ .$$
Of course we can as well compute the flux integral directly. To this end we parametrize $E_t$ as
$$E_t:\quad\theta\mapsto\sqrt{t}\bigl(\cos\theta,{1\over2}\sin\theta\bigr)\qquad(0\leq \theta\leq 2\pi)\ .$$
We then have
$${\bf n}\>{\rm d}s=(y_\theta, -x_\theta)\>d\theta=\sqrt{t}\bigl({1\over2}\cos\theta, \sin\theta\bigr)\>d\theta\ .$$
On the other hand $\nabla v=(2x,8y)$, so that our flux integral becomes
$$\int_{E_t}\nabla v\cdot {\bf n}\>{\rm d}s= \int_0^{2\pi}(\sqrt{t})^2(2\cos\theta,4\sin\theta)\cdot\bigl({1\over2}\cos\theta, \sin\theta\bigr)\>d\theta=5\pi t\ ,$$
as before.
Now we are interested in some other function $u(x,y):=\phi\bigl(v(x,y)\bigr)$ with a $\phi$ yet to be determined. The chain rule gives
$$\nabla u(x,y)=\phi'\bigl(v(x,y)\bigr)\>\nabla v(x,y)\ ,$$
whereby $\phi'(t)$ is a scalar factor. Since $v(x,y)\equiv t$ on $E_t$ we therefore have
$$\int_{E_t}\nabla u\cdot{\bf n}\>{\rm d}s=\phi'(t)\int_{E_t}\nabla v\cdot{\bf n}\>{\rm d}s=5\pi\>t\>\phi'(t)\ .$$
The right hand side is independent of $t$ iff $\phi'(t)={C\over t}$ for some constant $C$, and this is equivalent with
$$\phi(t)=C\>\log t+ C'$$
with $C'$ constant as well.
