How to prove this formula for surface integrals?

Question

Let $f(x)$ be a continuous function on $\mathbb{R}$. Prove $$\iint\limits_{S}f(x+y+z)dS=2\pi\int_{-1}^{1}f(\sqrt{3}\xi) d\xi$$ where S is $x^2+y^2+z^2=1$. I have already noticed that $\iint\limits_{S}f(x+y+z)dS=\iint\limits_{S}f(x+y+z)\cdot \frac{-1}{z}dx\wedge dy$, then I tried to let $$\left\{\begin{matrix} x=&\cos{\varphi} \cdot \cos{\phi} \\ y=&\cos{\varphi} \cdot \sin{\phi} \\ z=&\sin{\varphi} \end{matrix}\right.$$ so the formula turns into $\iint\limits_{\varphi\in[0,2\pi],\phi\in[-\frac{\pi}{2},\frac{\pi}{2}]}f(\cos{\varphi} \cdot \cos{\phi}+\cos{\varphi} \cdot \sin{\phi}+\sin{\varphi})\cdot \cos {\varphi}d\varphi d\phi$ But then I was stucked at here. I want to use Fubini Theorem but I can't separate $\cos{\varphi}$ out of the formula. How can I do next ?

Answer 1

Basically we are going to use spherical coordinates with the $z$-axis along the line $\{t(1,1,1):t\in\mathbb{R}\}$.

We first apply the rotation of $45^{\circ}$ around $z$-axis: $$\begin{cases} x=\frac{x'+y'}{\sqrt{2}}\\ y=\frac{x'-y'}{\sqrt{2}}\\ z=z' \end{cases}.$$ Hence $$I:=\iint\limits_{S}f(x+y+z)dS=\iint\limits_{S}f(\sqrt{2}x'+z')dS.$$ Then we apply the rotation of $\alpha^{\circ}$ around $y'$-axis where $\alpha^{\circ}$ is the angle between the vector $(1,1,1)$ and the plane $z=0$ (note that $\cos(\alpha^{\circ})=\frac{(1,1,1)}{\sqrt{3}}\cdot(0,1,0)=\frac{1}{\sqrt{3}}$): $$\begin{cases} x'=\frac{x''+\sqrt{2}z''}{\sqrt{3}}\\ y'=y''\\z'=\frac{-\sqrt{2}x''+z''}{\sqrt{3}} \end{cases}.$$ Hence $$I=\iint\limits_{S}f\Big(\sqrt{2}\Big(\frac{x''+\sqrt{2}z''}{\sqrt{3}}\Big)+\Big(\frac{-\sqrt{2}x''+z''}{\sqrt{3}}\Big)\Big)dS =\iint\limits_{S}f(\sqrt{3}z'')dS.$$ Note that $S$ is invariant with respect of those two rotations.

Finally we use spherical coordinates with the new $z''$-axis : $$I=\iint\limits_{S}f(\sqrt{3}z'')dS=\int_{\theta=0}^{2\pi}\int_{\phi=-\pi/2}^{\pi/2}f(\sqrt{3}\sin(\phi))\cos(\phi)d\theta d\phi=2\pi\int_{-1}^{1}f(\sqrt{3}\xi) d\xi$$ where $\xi=\sin(\phi)$.