Change of variables (proof of Morrey's inequality) I have one question of one of the steps in the proof of Morrey's inequality in Evans second edition. 
That is in the changing of variable, with $y = x+tw$, $t = |x-y|$. Evans do the following:
$$ \int_0^s \int_{ \partial B(0,1) } |Du(x+tw)|dS(w) dt = \int_0^s \int_{\partial B(x,t)} \frac{|Du(y)|}{t^{n-1}}dS(y)dt.$$
Here $x,y,w \in \mathbb{R}^n$.I'm wondering how the $t^{n-1}$ appears in the denominator? There should be a jacobian, due to the change of variable, however I am unsure how this is treated in the change from $dS(w)$ to $dS(y)$. 

Hausdorff Measure and The Coarea formula are what you need to make it really rigorous. The  Coarea formula though is in the appendix and working through Evans PDE book I reckon you can take it as a sort of 'black box' result. As a reference though have a look at sections 2 and 3 of Gariepy and Evans book: Measure Theory and Fine Properties of Functions.
In Evans's notation in the surface integral over an n-dim ball in $\mathbb{R}^n$ the $dS$ corresponds to the Hausdorff $n-1$ dim measure of the surface of the ball. In other words, you get the surface area of the $n$ dim ball. In particular,
\begin{equation}
\mathcal{H}^{n-1}(\partial B(x, r))=\int_{\partial B(x, r)}\mathrm{d}S=n\alpha(n)r^{n-1}.
\end{equation}
So I guess one way to 'see' the change of variables which is just a scaling map in your work above is as follows. We have
\begin{equation}
\int_{0}^{s}\int_{\partial B(0, 1)}\vert Du(x+tw)\vert\mathrm{d}S\mathrm{d}t= \int_{0}^{s}\int_{\partial B(0, 1)}\vert Du(x+tw)\vert\frac{t^{n-1}}{t^{n-1}}\mathrm{d}S\mathrm{d}t.
\end{equation}
Let $y=x+tw$, then
\begin{equation}
t=\vert x-y\vert\Rightarrow y\in\partial B(x, t)
\end{equation}
and (symbolically)
\begin{equation}
dS_y=n\alpha(n)t^{n-1}=t^{n-1}dS
\end{equation}
because
\begin{equation}
\mathcal{H}^{n-1}(\partial B(0, 1))=\int_{\partial B(0, 1)}\mathrm{d}S=n\alpha(n).
\end{equation}
So now putting everything together we have
\begin{equation}
\int_{0}^{s}\int_{\partial B(0, 1)}\vert Du(x+tw)\frac{t^{n-1}}{t^{n-1}}\mathrm{d}S\mathrm{d}t=\int_{0}^{s}\int_{\partial B(x, t)}\vert\frac{\vert Du(y)\vert}{t^{n-1}}\mathrm{d}S_y\mathrm{d}t
\end{equation}
