Suppose $0$ is an interior point of the convex domain $D\sbe\R^d$. Show that
$$
\frac{\vol{}(D)}{\vol{}(B_2^d)}=\int_{S^{d-1}}p_D(z)^{-d}\,\s(dz)~.
$$
where $B_2^d$ denotes the euclidean unit ball and $p_D(z)\colon=\inf\{r > 0: z\in rD\}$ is the so called Minkowski functional of $D$, i.e. $D=D^\circ=[p_D < 1]$ and $\pa D=[p_D=1]$.
Let $t_2(z)$ be the euclidean distance of the point of intersection of the boundary $\pa D$ and the half line $\R^+z$. By integration in 'polar coordinates':
\begin{eqnarray*}
\vol{}(D)
&=&\vol{}(S^{d-1})\int_{S^{d-1}}\int_0^{t_2(z)}t^{d-1}\,dt\,\s(dz)\\
&=&\vol{}(S^{d-1})d^{-1}\int_{S^{d-1}}t_2(z)^d\,\s(dz)
=\vol{}(B_2^d)\int_{S^{d-1}}t_2(z)^d\,\s(dz)~.
\end{eqnarray*}
As $t_2(z)z$ lies on the boundary, we have: $1=p_D(t_2(z)z)=t_2(z)p_D(z)$.