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

Question

Show that $$\iint_F f(ax+by+cz)dS = 2 \pi \int_{-1}^1 f(u \sqrt{a^2+b^2+c^2})du$$ where $F$ is the sphere $x^2 + y^2 +z^2 = 1$

Please provide me hints on how to proceed with this proof.

Answer 1

If you let $\vec n=(a,b,c)$ then the integrand function can be written as $f(ax+by+cz)=f(\vec n\cdot\vec r)$. Choose the reference frame so that $\vec n$ lies along the $z$ axis: it follows that $f(\vec n\cdot\vec r)=f(nz)$, where $n=\sqrt{a^2+b^2+c^2}$. Integrate then in spherical coordinates: $$ \int \int_F f(ax+by+cz)dS = \int_{0}^{2\pi}d\phi\int_0^\pi f(n\cos\theta)d(-\cos\theta)={2\pi}\int_{-1}^1 f(nu)du. $$