solution of wave equations in odd dimension Evans PDE Here I am looking at the proof of theorem 2 below

Here I have the following difficulties: 
1) In the last two lines, the exponent changes from $\frac{n-1}{2}$ to $\frac{n-3}{2}$, why? Could anyone explain?
2) To show (iii), i.e. check I.C. satisfied. I follow identities in lemma2 (ii) and (iii).(See picture below)
 But it seems I am lost and have messed up my calculation again. Could anyone help?

 A: The set on which the integral is evaluated also changes.  Here he's using polar coordinates to make the calculation work:
$$
\int_{B(x,t)} f(y) dy = \int_0^t \int_{\partial B(x,s)} f(z) d\sigma(z)   ds
$$
where $d\sigma$ is the surface measure on the sphere.  Hence the fundamental theorem of calculus says that
$$
\frac{d}{dt} \int_{B(x,t)} f(y) dy = \int_{\partial B(x,t)} f(z) d\sigma(z). 
$$
In the book he is taking one copy of the operator $t^{-1} \frac{d}{dt}$ and applying it as above, which then decreases the power by one and results in the term
$$
\frac{1}{t} \int_{\partial B(x,t)} \Delta h .
$$
A: For one, $$\left(\frac{1}{t}\frac{d}{dt}\right)^{\frac{n-1}{2}} = \left(\frac{1}{t}\frac{d}{dt}\right)^{\frac{n-3}{2}}\left(\frac{1}{t}\frac{d}{dt}\right)$$ Then note that the quantity $\left(\frac{1}{t}\frac{d}{dt}\right)$ appears to get tucked into the integral. This is why on the second line in question you have a $\frac{1}{t}$ outside the integral. The infinitesimal also changes from $dy$ to $dS$ along with the integral region. Using the formula $$\frac{d}{dt}\left(\int_{B(x_0,t)}f dx\right)=\int_{\partial B(x_0,t)} f dS$$ you mentioned as appearing on appendix pg. 628 of Evans, the result follows. 
A: According to graydad's answer, I have 

with use of 
 
