$\int_{0\leq v\leq 2\pi,-1\leq t\leq 1}f(at+\sqrt{1-t^2}(b\cos v+c\sin v))dtdv=2\pi\int_{-1}^{1}f(u\sqrt{a^2+b^2+c^2})du$

Question

Suppose $a, b, c$ are given real numbers, such that all of them are not zero.

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous function on $[-\sqrt{{a}^{2}+{b}^{2}+{c}^{2}},\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}]$, prove that $$\int_{0\leq v\leq 2\pi,-1\leq t\leq 1}f(at+\sqrt{1-t^2}(b\cos v+c\sin v))dtdv=2\pi\int_{-1}^{1}f(u\sqrt{a^2+b^2+c^2})du$$

I have noticed that $$LHS=\int_{0\leq v\leq 2\pi,-1\leq t\leq 1}f(at+\sqrt{1-t^2}\sqrt{{b}^{2}+{c}^{2}}\mathrm{sin}(v+\alpha ))dtdv$$ $$={\int }_{-1}^{1}({\int }_{0}^{2\pi }f(at+\sqrt{1-t^2}\sqrt{{b}^{2}+{c}^{2}}\mathrm{sin}\left(v+\alpha \right))dv)dt$$ (By Fubini's theorem) $$={\int }_{-1}^{1}({\int }_{0}^{2\pi }f(at+\sqrt{1-t^2}\sqrt{{b}^{2}+{c}^{2}}\mathrm{sin}v)dv)dt$$ (Since $f(x+2\pi)=f(x)$) But then what can be done to complete the proof?

Answer 1

Letting $t=\cos(u)$ then the given integral is a surface integral over the unit sphere: $$\int_{v=0}^{2\pi}\int_{u=0}^{\pi}f(a\cos(u)+\sin(u)(b\cos v+c\sin v))\sin(u) dudv=\iint_S f(ax+by+cz)\,dS.$$ where $x=\cos(u)$, $y=\sin(u)\cos(v)$, $z=\sin(u)\sin(v)$, and $S=\{(x,y,z): x^2+y^2+z^2=1\}$.

Finally, apply Show that $\iint_F f(ax+by+cz)dS = 2 \pi \int_{-1}^1 f(u \sqrt{a^2+b^2+c^2})du$

In a few words, the integrals on both sides are surface integral over the unit sphere expressed in spherical coordinates, but on the right-side the axes are conveniently chosen.