Evaluate the volume between a parabola and a sphere in high dimension

Question

Consider the unit ball $\mathbb{B}^{n+1}:=\{\boldsymbol{x}\in\mathbb{R}^{n+1}:\|\boldsymbol{x}\|_2\le 1\}$ and a parabola $p:\mathbb{R}^{n}\rightarrow\mathbb{R}$ for which $p(\boldsymbol{x}):=\rho\|\boldsymbol{x}\|_2^2/2$, where $\rho>0$ is some parameter. I would like to know how the volume of $\mathbb{B}^{n+1}\cap\operatorname{epi}(p)=\mathbb{B}^{n+1}\cap\{(\boldsymbol{x},t)\in\mathbb{R}^{n}\times\mathbb{R}:p(\boldsymbol{x})\le t\}$ scales w.r.t. $\rho$. (Notice that when $\rho=0$, the interested volume is exactly the half of that of $\mathbb{B}^{n+1}$). Thanks in advance.

Answer 1

\begin{align} \mathcal V\left(\mathbb B^{n+1} \cap \textbf{epi}(p)\right) &= \int_{-1}^{1} \left(\int_{\left\|\boldsymbol x\right\|^2 \le 1-t^2 \cap \left\|\boldsymbol x\right\|^2 \le \frac2\rho t}\mathrm d \boldsymbol x\right)\mathrm d t\\ &= \int_{0}^{1} \mathcal V\left(\mathbb B^{n}\right) \left(\min \left\{1-t^2, \frac2\rho t\right\}\right)^n\mathrm d t\\ &= \mathcal V\left(\mathbb B^{n}\right)\left(\left(\frac2{\rho}\right)^n\int_0^{\phi} t^n \mathrm dt + \int_{\phi}^{1} \left(1-t^2\right)^n \mathrm d t\right) \end{align}

where $\phi > 0$ such that $1 - t^2 - \frac{2}{\rho}t = 0$. This implies that $$\phi = -\frac1\rho + \sqrt{\frac1{\rho^2} + 1} < 1.$$

Let \begin{align} \mathbf{I}_n\left(\phi\right) &= \int_{\phi}^{1} \left(1-t^2\right)^n \mathrm d t \end{align}

Since $\mathcal V\left(\mathbb B^n\right) = \displaystyle\frac{\pi^{\frac{n}2}}{\Gamma\left(\frac{n}2 + 1\right)}$

So

\begin{align} \mathcal V\left(\mathbb B^{n+1} \cap \textbf{epi}(p)\right) &= \displaystyle\frac{\pi^{\frac{n}2}}{\Gamma\left(\frac{n}2 + 1\right)}\left(\left(\frac2{\rho}\right)^n\frac{\phi^{n+1}}{n+1} + \mathbf{I}_n\left(\phi\right) \right)\\ &= \displaystyle\frac{\pi^{\frac{n}2}}{\Gamma\left(\frac{n}2 + 1\right)}\left(\frac{1}{n+1} \phi\left(1-\phi^2\right)^{n} + \mathbf{I}_n\left(\phi\right)\right) \end{align}

Let, $\mathbf {J}_n (\phi) = \frac1{n+1} \phi \left(1 - \phi^2\right)^{n} + \mathbf{I}_n(\phi)$

\begin{align} \frac{\mathrm d \mathbf{J}_n}{\mathrm d \phi} &= \frac1{n+1} \left(1-\phi^2\right)^{n} - \frac{2n}{n+1}\phi^2\left(1-\phi^2\right)^{n-1} - \left(1-\phi^2\right)^{n} < 0. \end{align}