referido computo, que resulta de uma discussao ocorrida em 2013 no forum Mathematics Stack Exchange.Solucao do Problema: Para $x\in\mathbb{R}^n$ e $t>0$, temos$$\Phi(x,t)=\frac{1}{(4\pi t)^{n/2}}\mathrm{e}^{\frac{-|x|^2}{4t}}$$Portanto, para $y\in\mathbb{R}^n$ e $s<0$,$$\begin{align*}\Phi(-y,-s)\geq 1\quad\Longleftrightarrow& \quad\frac{1}{(-4\pi s)^{n/2}}\mathrm{e}^{\frac{|y|^2}{4s}}\geq1\\\\\Longleftrightarrow& \quad(-4\pi s)^{n/2}\leq\mathrm{e}^{\frac{|y|^2}{4s}}\\\\\Longrightarrow& \quad(-4\pi s)^{n/2}\leq 1\\\\\Longleftrightarrow& \quad s\geq -\frac{1}{4\pi}\end{align*}$$e$$\begin{align*}\Phi(-y,-s)\geq 1\quad\Longleftrightarrow& \quad\ln\left(\frac{1}{(-4\pi s)^{n/2}}\mathrm{e}^{\frac{|y|^2}{4s}}\right)\geq0\\\\\Longleftrightarrow& \quad(-n/2)\ln\left(-4\pi s\right)+\frac{|y|^2}{4s}\geq0\\\\\Longleftrightarrow& \quad|y|^2\leq 2ns\ln(-4\pi s)\end{align*}$$Isto implica que$$\begin{align*}E(0,0;1)&=\{(y,s)\in\mathbb{R}^{n+1}\mid s\leq 0,\;\Phi(-y,-s)\geq 1\}\\\\&=\left\{(y,s)\in\mathbb{R}^{n+1}\;\Big|\; -\frac{1}{4\pi}\leq s\leq 0,\;|y|^2\leq 2ns\ln(-4\pi s)\right\}\end{align*}$$Assim,$$\begin{align*}\iint_{E(0,0;1)}\frac{|y|^2}{s^2}\;dyds&=\int_{-\frac{1}{4\pi}}^0\int_{|y|^2\leq2ns\ln(-4\pi s)}\frac{|y|^2}{s^2}\;dyds\\\\&=\int_{-\frac{1}{4\pi}}^0\int_0^{\sqrt{2ns\ln(-4\pi s)}}\int_{\partial B(0,r)}\frac{|y|^2}{s^2}\;dS(y)drds \quad\text{(coordenadas polares)}\\\\&=\int_{-\frac{1}{4\pi}}^0\int_0^{\sqrt{2ns\ln(-4\pi s)}}\int_{\partial B(0,1)}\frac{|rw|^2}{s^2}r^{n-1}\;dS(w)drds \quad\text{(mudanca $w=y/r$)}\\\\&=\int_{-\frac{1}{4\pi}}^0\int_0^{\sqrt{2ns\ln(-4\pi s)}}\int_{\partial B(0,1)}\frac{r^{n+1}}{s^2}\;dS(w)drds\\\\&=\text{med}\big(\partial B(0,1)\big)\int_{-\frac{1}{4\pi}}^0\int_0^{\sqrt{2ns\ln(-4\pi s)}}\frac{r^{n+1}}{s^2}drds\\\\&=\text{med}\big(\partial B(0,1)\big)\int_{-\frac{1}{4\pi}}^0\frac{(2ns\ln(-4\pi s))^{(n+2)/2}}{(n+2)s^2}ds\\\\&=\frac{(2n)^{(n+2)/2}}{(n+2)}\text{med}\big(\partial B(0,1)\big)\int_{-\frac{1}{4\pi}}^0\frac{(s\ln(-4\pi s))^{(n+2)/2}}{s^2}ds\end{align*}$$Calculando esta ultima integral, obtemos$$\begin{align*}\int_{-\frac{1}{4\pi}}^0\frac{(s\ln(-4\pi s))^{(n+2)/2}}{s^2}ds&=\int_{1}^0\frac{\left((-\frac{x}{4\pi})\ln(x)\right)^{(n+2)/2}}{(-\frac{x}{4\pi})^2}\left(-\frac{1}{4\pi}\right)dx\quad\text{(mudanca $x=-4\pi s$)}\\\\&=\int_{0}^\infty\frac{\left(\frac{\mathrm{e}^{-z}}{4\pi}z\right)^{(n+2)/2}}{(-\frac{\mathrm{e}^{-z}}{4\pi})^2}\left(-\frac{1}{4\pi}\right)(-\mathrm{e}^{-z})\;dz\quad\text{(mudanca $z=-\ln(x)$)}\\\\&=\frac{1}{(4\pi)^{n/2}}\int_{0}^\infty\mathrm{e}^{-nz/2}z^{(n+2)/2}\;dz\\\\&=\frac{1}{(4\pi)^{n/2}}\int_{0}^\infty\mathrm{e}^{-w}\left(\frac{2w}{n}\right)^{(n+2)/2}\frac{2}{n}\;dw\quad\text{(mudanca $w=nz/2$)}\\\\&=\frac{1}{(4\pi)^{n/2}}\left(\frac{2}{n}\right)^{n/2+2}\int_{0}^\infty\mathrm{e}^{-w}w^{n/2+1}\;dw\\\\&=\frac{1}{2^{n}\pi^{n/2}2^{-n/2-2}n^{n/2+2}}\int_{0}^\infty\mathrm{e}^{-w}w^{n/2+2-1}\;dw\\\\&=\frac{1}{\pi^{n/2}2^{n/2-2}n^{n/2+2}}\Gamma\left(2+\frac{n}{2}\right)\quad\text{(definicao da funcao gama)}\end{align*}$$Deste modo,$$\begin{align*}\iint_{E(0,0;1)}\frac{|y|^2}{s^2}\;dyds&=\frac{(2n)^{(n+2)/2}}{(n+2)}\text{med}\big(\partial B(0,1)\big)\frac{1}{\pi^{n/2}2^{n/2-2}n^{n/2+2}}\Gamma\left(2+\frac{n}{2}\right)\\\\&=\frac{8\;\text{med}\big(\partial B(0,1)\big)}{n(n+2)\pi^{n/2}}\Gamma\left(2+\frac{n}{2}\right)\\\\&=\frac{8\;\text{med}\big(\partial B(0,1)\big)}{n(n+2)\pi^{n/2}}\left(1+\frac{n}{2}\right)\Gamma\left(1+\frac{n}{2}\right)\quad\text{(propriedade $\Gamma(t+1)=t\Gamma(t)$)}\\\\&=\frac{4\;\text{med}\big(\partial B(0,1)\big)}{n\pi^{n/2}}\Gamma\left(1+\frac{n}{2}\right)\\\\&=\frac{4\;\text{med}\big(\partial B(0,1)\big)}{n\pi^{n/2}}\frac{n}{2}\Gamma\left(\frac{n}{2}\right)\quad\text{(mesma propriedade anterior)}\\\\&=4\ \text{med}\big(\partial B(0,1)\big)\frac{\Gamma\left(\tfrac{n}{2}\right)}{2\pi^{n/2}}\\\\&=4\quad \text{(propriedade med$(\partial B(0,1))=\tfrac{2\pi^{n/2}}{\Gamma(n/2)}$)}\end{align*}$$ Nota: A definicao e as propriedades utilizadas da funcao gama podem ser encontradas nas paginas 7 e 8 da segunda edicao do livro Introduction to partial differential equations de G. B. Folland.Referencias: aqui e aqui.