Transformation of domain in Evans From Evans, Partial Differential Equations, Page 53. 
Let $\Phi(x,s)=\frac{1}{{4\pi t}^{n/2}}e^{-\frac{|x|^{2}}{4t}}$. Evans used $E(x,t,r)$ to denote the region $$(y,s)\in \mathbb{R}^{n+1}|s\le t, \Phi(x-y,t-s)\ge \frac{1}{r^{n}}$$
In particular he used $E(r)$ to denote the region $E(0,0,r):\Phi(-y,-s)\ge \frac{1}{r^{n}}$. 
My question is how the following two equalities (here $u$ is an $C^{2}$ function in $y$ and $s$):
$$\int\int_{E(1)}\sum^{n}_{i=1}u_{y_{i}}y_{i}\frac{|y|}{s^{2}}+2ru_{s}\frac{|y|^{2}}{s}dyds=\frac{1}{r^{n+1}}\int\int_{E(r)}\sum^{n}_{i=1}u_{y_{i}}y_{i}\frac{|y|}{s^{2}}+2u_{s}\frac{|y|^{2}}{s}dyds$$
and 
$$\frac{1}{r^{n}}\int\int_{E(r)}u(y,s)\frac{|y|^{2}}{s^{2}}dyds=\int\int_{E(1)}u(ry,r^{2}s)\frac{|y|^{2}}{s^{2}}dyds$$
The transformation of domain is the obvious issue at here. For any function $u(y,s)$, to transform $\int\int_{E(r)}u(y,s)$ to $\int\int_{E(1)}u(y',s')$ one need to make certain dilation and change of variables. The reason is when we change $\Phi(x,t)\rightarrow \Phi(rx,r^{2}t)$, there is an extra $\frac{1}{r^{n}}$ term such that $\frac{1}{r^{n}}\Phi(x,t)=\Phi(rx,r^{2}t)$. Therefore $\Phi(-y,-s)\ge 1$ implies $\Phi(-ry,-r^{2}s)\ge \frac{1}{r^{n}}$. So $(y,s)\rightarrow (ry,r^{2}s)$ change $E(1)$ to $E(r)$. 
But in the second inequality the situation is reversed. I do not really know how to reach from here to the above equalities, so I decided to ask. 
 A: I get easily confused when people change the variables of integration but keep the same names. 
Especially when derivatives get involved as well. (For example, what does $u_x(2x,y)$ mean?)
Here is my preferred notation: use $(y,s)$ for points in $E(r)$ and $(z,t)$ for points in $E(1)$. They are related by 
$$
y=rz, \ s=r^2t, \quad dy\,ds =r^{n+2}\,dz\,dt 
\tag1$$
Also, when the derivatives get involved, I will use numeric subscripts instead of letters. That is, I will write $u_i$ for the 
derivative of $i$ with respect to its $i$th argument. The arguments will be enumerated as: $s,y_1,\dots,y_n$, with $s$ being the $0$th argument. 
Now that were are armed with proper notation, let's attack:
$$
\begin{split}
\phi(r)&=r^{-n}\iint_{E(r)} u(y,s)\,\frac{|y|^2}{s^2}\,dy\,ds  \\
& =  r^{-n}\iint_{E(1)} u(rz,r^2t)\,\frac{r^2|z|^2}{r^4 t^2}\,r^{n+2}\,dz\,dt \\
& =  \iint_{E(1)} u(rz,r^2t)\,\frac{|z|^2}{t^2}\,dz\,dt 
\end{split}\tag2 $$
The derivative of $u(rz,r^2t)$ with respect to $r$ is $2rtu_0(rz,r^2t)+\sum_{i=1}^n z_i u_i(rz,r^2t)$. Plug this in and return
to the original variables: 
$$
\begin{split}
& \iint_{E(1)} \left(2rtu_0(rz,r^2t)+\sum_{i=1}^n z_i u_i(rz,r^2t)\right)\,\frac{|z|^2}{t^2}\,dz\,dt   \\
& = \iint_{E(1)} \left(2rr^{-2}s u_0(y,s)+\sum_{i=1}^n r^{-1}y_i u_i(y,s)\right)\,\frac{r^{-2}|y|^2}{r^{-4}s^2}r^{-n-2}\,dy\,ds  \\
& = r^{-n-1}\iint_{E(1)} \left(2s u_0(y,s)+\sum_{i=1}^n y_i u_i(y,s)\right)\,\frac{|y|^2}{s^2}\,dy\,ds
\end{split}\tag3 $$
which matches the formula in the book.
