I hope everyone is going well today.
Is anyone know how to compute the Lebesgue measure of the red cone below ?
Here's more details, I take the ball $B(0,r)$ and the point $q \notin B(0,r)$. Then I take the intersection of this ball with the hyperplan passing through the origin and orthogonal to $q$ that is : $\Sigma_{r} = \{ z \in \mathbb{R}^{n} ; \langle z, q \rangle = 0 \} \cap B(0,r)$.
Let $C$ be the cone generated by $q$ and $\Sigma_{r}$ we have : \begin{equation} \mathcal{L}(C) = c_{n}|q|r^{n-1} \end{equation}
With $c_{n}>0$ a dimensional constant and $\mathcal{L}$ the Lebesgue measure. How to prove that ?
I wish you a very good day.
This is pretty obvious. Let $|q|=:h>0$. Then you may assume $q=(0,\ldots,0,h)$ and $\Sigma_r$ being the ball of radius $r$ in the $x'=(x_1,\ldots, x_{n-1})$-plane. The $(n-1)$-volume of this ball is $\kappa_{n-1} r^{n-1}$, where $\kappa_{n-1}$ is the volume of the unit $(n-1)$-ball. The cone $C$ then is a "right circular" cone of height $h$ standing on $\Sigma_r$. A plane $x_n=z\in[0,h]$ intersects $C$ in an $(n-1)$-ball $\Sigma_{\rho(z)}$ of radius $$\rho(z)=r\left(1-{z\over h}\right)\ .$$ Fubini's theorem then gives $${\rm vol}(C)=\int_C 1\>{\rm d}(x)=\int_0^h\int_{\Sigma_\rho(z)}1\>{\rm d}(x')\>dz=\int_0^h\kappa_{n-1}\rho^{n-1}(z)\>dz=\ldots={\kappa_{n-1}\over n}\>h\>r^{n-1}\ .$$