How to prove that $\iint\frac{|y|^2}{s^2}\,dy\,ds=4$? Let $U\subset\mathbb{R}^n$ be an open set, $\Phi$ the fundamental solution of heat equation, $T>0$, $r>0$, $x\in\mathbb{R}^n$ and $t\in\mathbb{R}$. Defines $U_T=U\times(0,T]$ and $$E(x,t;r)=\{(y,s)\in\mathbb{R}^{n+1};\:s\leq t,\;\Phi(x-y,t-s)\geq r^{-n}\}.$$
Evans PDE book presents the following 

Theorem (mean-value property for the heat equation): Let $u\in C^2_1(U_T)$ solve the heat equation. Then $$u(x,t)=\frac{1}{4r^n}\iint_{E(x,t;r)}u(y,s)\frac{|x-y|^2}{(t-s)^2}\,dy\,ds.$$ 

In the proof is used the following equality:
$$\iint_{E(0,0;1)}\frac{1}{s^2}\sum_{i=1}^n {y_i}^2\,dy\,ds=4$$ 
Could someone help me with details of this calculation?
Thanks.
 A: First of all, let's try to understand what is $E(0,0;1)$. Remember that $$\Phi(x,t)=\frac{1}{(4\pi t)^{n/2}}e^{\frac{-|x|^2}{4t}},\ \forall\ x\in\mathbb{R}^n,\ \forall\ t>0$$
The function $e^{\frac{-|x|^2}{4t}}$ has range in $(0,1]$, hence, if $\Phi(-y,-t)\geq 1$, we can conclude that $(4\pi (-t))^{n/2}\leq 1$, which implies that $t\in \left(\frac{-1}{4\pi},0\right)$.
Moreover, applying $\log$ in both sides of the equation $\Phi(-y,-t)\geq 1$, we conclude that $$|y|^2\leq 2ns\log{(4\pi (-s))},\ \forall\ s\in \left(\frac{-1}{4\pi},0\right)\tag{2} $$
Note that $(2)$ is the domain of integration. We have that 
$$
 \int_{-1/4\pi}^0\int_{|y|^2\leq 2ns\log{4\pi (-s)}} \frac{|y|^2}{s^2}dyds= \int_{-1/4\pi}^0\int_{0}^{\sqrt{2ns\log{(4\pi (-s))}}}\frac{r^2}{s^2} r^{n-1}drds\int_{S(0,1)}d\omega  $$
where $y=r\omega$ with $\omega\in S(0,1)=\{x\in\mathbb{R}^n:\ |x|=1\}$ (note that $dy=r^{n-1}drd\omega$). The last integral I calculated in Mathematica and got the result $4$. 
Remark: (see here) $$\int_{S(0,1)}d\omega=\frac{2\pi^{n/2}}{\Gamma(n/2)}$$
Remark 1: This post may be of some help.
