An equality in Evans' PDE book chapter 2 I don't know how to type the integral sign with a bar, so in the following, all of integral signs mean integral sign with a bar, i.e. average integral.
Let $x\in\mathbb R^n,t\gt 0,r\gt 0$.
Evans defines a function $U(x;t,r):=\int_{\partial B(x,r)}u(y,t)dS(y),$ where $u$ is a good function, which means it can be differentiated as many times as you want.(To avoid tedious background, I define $u$ in this way. The condition in Evans' book is weaker.)
Then $$U_r(x;r,t)=\frac{r}{n}\int_{B(x,r)}\Delta u(y,t)dy$$ by using the divergence theorem and change of variable.
And Evans omitted the proof for $$U_{rr}(x;r,t)=\int_{\partial B(x,r)}\Delta u\,dS+(\frac{1}{n}-1)\int_{B(x,r)}\Delta u\,dy.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)$$
I tried to obtain the equality $(1)$ but I failed. There is subtle difference between differentiation of $U$ and $U_r$, since the integral domains are different. And the trick for differentiating $U$ seems failing in differentiating $U_r$.
Can someone help me? Thank a lot in advance.

I will assume no bars on any of my integrals, and I will write $\omega_n = |B(0,1)|$ for the volume of the unit ball in $\mathbb{R}^n$.  Then from the first step Evans computes
$$
U_r(r,t) = \frac{r}{n \omega_n r^n} \int_{B(x,r)} \Delta u(y,t) dy =  \frac{1}{n \omega_n r^{n-1}} \int_{B(x,r)} \Delta u(y,t) dy.
$$
Using hyperspherical coordinates, we can write 
$$
\int_{B(x,r)} f(y) dy = \int_0^r \int_{\partial B(x,s)} f(z) d\sigma(z) ds
$$
where $d\sigma$ is surface measure on the sphere.  Thus 
$$
U_r(r,t) = \frac{1}{n \omega_n r^{n-1}} \int_0^r \int_{\partial B(x,s)} \Delta u(z,t) d \sigma(z) ds,
$$
and so we differentiate w.r.t. $r$ to get
$$
U_{rr}(r,t) = \frac{(1-n)}{n \omega_n r^{n}} \int_0^r \int_{\partial B(x,s)} \Delta u(z,t) d \sigma(z) ds + \frac{1}{n \omega_n r^{n-1}}  \int_{\partial B(x,r)} \Delta u(z,t) d \sigma(z) \\
=  \frac{(1-n)}{n \omega_n r^{n}} \int_{B(x,r)} \Delta u(y,t) dy + \frac{1}{n \omega_n r^{n-1}}  \int_{\partial B(x,r)} \Delta u(z,t) d \sigma(z).
$$
Taking into account the Hausdorff measure of the sphere, this is exactly the formula Evans gives.
