The function $\phi:L\to\mathbb{R}$ where $L={\{(x,y)\in\mathbb{R}^2:x^2+y^2=4\}}$ is defined as,
\begin{align*}&\phi(x,y)=\\ &\int_{0}^{\pi}\!\!\!\!\int_{0}^{2\pi}\!\!\!\!\int_{1}^{2}\!\!\frac{\cos^2(\Psi) \sin(\theta)}{r^2\sqrt{(x-r\cos(\Psi)\sin(\theta))^2+(y-r\sin(\Psi) \sin(\theta))^2+(r\cos(\theta))^2}}\,dr\,d\Psi\,d\theta \end{align*} for all $(x,y)\in L$. How can we show that $\phi$ is a constant or not on $L?$ It's better if numerical approximations can be found for maximum and minimum of the function $\phi$ on $L$.
The following may be useful:
$$\frac{d\phi}{d\alpha}=\iiint\frac{\cos^2(\Psi) \sin^2(\theta) \sin(\Psi-\alpha)}{r [{4+r^2-4r\sin(\theta)\cos (\Psi-\alpha)}]^{3/2}}\,dr\,d\Psi\,d\theta$$ when $(x,y)=(2\cos(\alpha),2\sin(\alpha))$
I found used numerical approximations at some points on $L$ using Mathematica numerical integrals to show that $d\phi/d\alpha\neq0$. Seems like difficult to use a analytical method. (Not a proof)