Prove: $\frac{\rho}{(4\pi)^2}\int_{|\xi|=1}d\omega_\xi\int_{|\eta|=1}f(x^0+r\xi+\rho\eta)d\omega_\eta=\int_{|r-\rho|}^{r+\rho}\frac{\lambda}{8\pi r}d\lambda\int_{|\zeta|=1}f(x^0+\lambda\zeta)d\omega_{\zeta}$,
where $d\omega_\eta$ is unit spherical area element. $r, \rho, \lambda >0$, $\xi, \eta, \zeta$ are all unit vectors in $\mathbb{R}^3$, $x_0 \in \mathbb{R}^3$(This is an equality in Fritz John's paper "blow-up of solutions of nonlinear wave equations in three dimensions")
I know we can calculate it by placing the region into a spherical shell, but I can't get the precisely form in the right and is there any other way to prove it?
Unfortunately I do not have the time to carefully write out all constants, so I only sketch the computation. Set $x_0=0$ without loss of generality. Next, denote $r\xi=x$, $\rho\eta=y$, $\lambda\zeta=z$. Clearly, as $x$ and $y$ vary over their spheres of radii $r$ and $\rho$, respectively, $z$ varies in the spherical layer $|r-\rho|\le z\le r+\rho$; for fixed $z$, both $r$ and $\rho$ vary on a circle (imagine a triangle with sides $\rho,r,|z|$; the circle is drawn by the vertex opposite to the side of length $|z|$ as we rotate the triangle around that side). Now it is clear that we should represent the integral on the LHS as an integral over that layer of the integral over the circle. We have $\omega_\xi=\operatorname{const} r^{-3}\Omega_x$, where $\Omega_x$ is the Leray form $$ \Omega_x=x_1dx_2\wedge dx_3+x_2dx_3\wedge dx_1+x_3dx_1\wedge dx_2. $$ Since $x_1^2+x_2^2+x_3^2=r$, we have $x_1dx_1+x_2dx_2+x_3dx_3=0$ and hence (in the chart where $x_1\ne0$) $$ \Omega_x=\frac r{x_1}dx_2\wedge dx_3. $$ Likewise, $$ \Omega_y=y_1dy_2\wedge dy_3+\dots=\frac\rho{y_1}dy_2\wedge dy_3. $$ If we take $z=x+y$, then routine computation shows that $$ \Omega_x\wedge\Omega_y=d^3z\wedge\alpha\equiv dz_1\wedge dz_2\wedge dz_3\wedge\alpha, $$ where $\alpha=a_1d(x_1-y_1)+a_2d(x_2-y_2)+a_3d(x_3-y_3)$ and the vector $a=(a_1,a_2,a_3)$ is determined by the condition that the triple product of $z$, $x$, and $a$ is $2\rho r$. Now (up to multiplicative constants) we can rewrite the integral on the LHS as $$ LHS=\int f(z)d^3z\wedge\alpha $$ and then compute the integral with respect to $\alpha$ over the above-mentioned circle depending on $z$. This will give the RHS. Hopefully you can reconstruct what I have omitted from the computation and restore all constants etc.