Computing the spherical mean and showing it satisfies PDE Compute the spherical mean of the function $h : \mathbb R^3 \to \mathbb R$ with
$$
 h(x,y,z) = x
$$
and show that it satisfies the differential equation
$$
 u_{rr} + \frac{2}{r} u_r = u_{xx} + u_{yy} + u_{zz}
$$
(note that the spherical mean of a function is a function of some fixed point and a radius).
 A: With $x\in\mathbb{R}^3$ and $r>0$, let us write
$$
M(x,r) = \frac1{4\pi}\int_{S^2}h(x+r\xi)\,\mathrm{d}^2\xi,
$$
where the integral is taken over the two dimensional unit sphere $S^2$.
Then the Laplacian of $M$ with respect to $x$ is
$$
\Delta_x M(x,r) = \frac1{4\pi}\int_{S^2} \Delta_x h(x+r\xi)\,\mathrm{d}^2\xi
= \frac1{4\pi}\int_{S^2} \Delta h(x+r\xi)\,\mathrm{d}^2\xi.
$$
Further, using the formula for the Laplacian in spherical coordinates, we have
$$
\Delta_x M(x,r) = \frac1{4\pi}\int_{S^2} (\partial_r^2+\frac2r\partial_r+\Delta_{S_r(x)}) h(x+r\xi)\,\mathrm{d}^2\xi ,
$$
where $\Delta_{S_r(x)}$ is the Laplace-Beltrami operator associated to the sphere of radius $r$ centred at $x$.
Finally, observing that the Laplace-Beltrami term integrates to zero, and taking the radial derivative operators outside the integral, we get the desired result. Note that all manipulations are justified if $h$ is a $C^2$ function in the region under consideration.
An alternative way, that avoids the explicit mention of the Laplace-Beltrami operator is to establish that the Laplace operator commutes with spherical mean, in the sense that 
$$
\int_{S^2} \Delta h(x+r\xi)\,\mathrm{d}^2\xi = \int_{S^2} \Delta g(x+r\xi)\,\mathrm{d}^2\xi,
$$
where
$$
g(y) = \frac1{4\pi}\int_{S^2} h(x+r\xi)\,\mathrm{d}^2\xi,
\qquad \textrm{for all}\quad y\in S_r(x).
$$
A: Inspired by epimorphic's comment, I decided to derive the formula to see what happens. I believe the question should be phrased as: say with slightly abuse of notation let us denote $\vec{x} = (x,y,z)$ and $r>0$ be a fixed radius, then

Let $u(\vec{x},r)$ be the spherical mean of $h:\mathbb{R}^3\to \mathbb{R}$:
  $$u(\vec{x},r) := \frac{1}{\omega_{2}(r)}\int_{\partial B(\vec{x},r)} h(\vec{y})\,dS(\vec{y}), \tag{1}$$
  the problem wants to investigate whether or not
  $$\frac{\partial^2}{\partial r^2} u(\vec{x},r) + \frac{2}{r} \frac{\partial}{\partial r} u(\vec{x},r) = \Delta_{\vec{x}}u(\vec{x},r):= \left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}\right)u(\vec{x},r).$$

Using the second formula the wikipedia entry of spherical mean provides:
$$u(\vec{x},r) = \frac{1}{\omega_{2}(1)}\int_{\|\vec{y}\|=1} \! h(\vec{x}+r\vec{y}) \, dS(\vec{y}).$$
This formula gives us an easier form to compute $\dfrac{\partial}{\partial r}$ than the first, we have
$$
\frac{\partial}{\partial r}u(\vec{x},r) = \frac{1}{\omega_{2}(1)}\int_{\|\vec{y}\|=1} \! \nabla_{\vec{x}} \,h(\vec{x}+r\vec{y}) \cdot \vec{y}\, dS(\vec{y}). 
$$
Also due to the unit sphere nature of the formula, we can exploit the divergence theorem on this formula: change of the derivative w.r.t. $\vec{y}$ and use the fact that $\vec{y}$ is the exterior unit vector normal to the unit sphere $\{\|y\|=1\}$.
$$
\frac{\partial}{\partial r}u(\vec{x},r) =\frac{1}{\omega_{2}(1)}\int_{\|\vec{y}\|=1} \! \frac{1}{r}\nabla_{\vec{y}} \,h(\vec{x}+r\vec{y}) \cdot \vec{y}\, dS(\vec{y})
\\
= \frac{1}{\omega_{2}(1)}\int_{\|\vec{y}\|<1} \! \frac{1}{r}\Delta_{\vec{y}} \,h(\vec{x}+r\vec{y})\,d\vec{y}
\\
= \frac{1}{\omega_{2}(1)}\int_{\|\vec{y}\|<1} \! r\Delta_{\vec{x}} \,h(\vec{x}+r\vec{y})\,d\vec{y}.
$$
Now use the spherical shell integration formula again in Evans's Appendix C.3 (see also his application in 2.2.2):
$$
\frac{\partial}{\partial r}u(\vec{x},r) =\frac{1}{\omega_{2}(1)}r\Delta_{\vec{x}} \frac{1}{r^3}
\int_{\|\vec{y}\|<r} h(\vec{x}+ \vec{y})\,d\vec{y}
\\
=\frac{1}{\omega_{2}(1)r^2}\Delta_{\vec{x}}
\int^r_0\left(\int_{\partial B(\vec{x},s)} h(\vec{y}')\,dS(\vec{y}')\right)ds.
$$
Comparing with (1) we have:
$$
\frac{\partial}{\partial r}u(\vec{x},r) 
=\frac{1}{r^2}\Delta_{\vec{x}}
\int^r_0 s^2\left(\frac{1}{\omega_{2}(1)s^2}\int_{\partial B(\vec{x},s)} h(\vec{y}')\,dS(\vec{y}')\right)ds
\\
= \frac{1}{r^2}\Delta_{\vec{x}}\int^r_0 s^2 u(\vec{x},s)\,ds. 
$$
This implies
$$
\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}u(\vec{x},r) \right)
= \Delta_{\vec{x}} (r^2 u(\vec{x},r)),
$$
which reduces to
$$
r^2\frac{\partial^2}{\partial r^2}u(\vec{x},r) + 2r \frac{\partial}{\partial r}u(\vec{x},r) = r^2 \Delta_{\vec{x}}  u(\vec{x},r),
$$
the same formula in the question above.
A: Here we assume that the center of the ball with radius $r$ is at $(r=0,\theta=0,\phi=0)$. The spherical mean of $h(x,y,z)=x$ is given by:
$$\frac{1}{4\pi r^2}\int_0^{\pi}rd\theta \int_0^{2\pi}r\sin\theta d\phi[r \sin\theta\cos\phi]=\frac{r^3}{4\pi r^2}\int_0^{\pi}\sin^2\theta d\theta \int_0^{2\pi} \cos\phi d\phi=0$$
Of course $u=0$ satisfies the homogeneous PDE:
$$u_{rr} + \frac{2}{r} u_r = u_{xx} + u_{yy} + u_{zz}$$
Now we assume that the center of the ball with radius $r$ is at $(x_0,y_0,z_0)$.
So that $\vec R=(x,y,z)=(x',y',z')+(x_0,y_0,z_0)$.  We now set $x'=r \sin\theta\cos\phi,y'=r \sin\theta\sin\phi,z'=r \cos\theta$.
The spherical mean of $h(x,y,z)=x=x_0+x'=x_0+r \sin\theta\cos\phi$ is given by $I_1+I_2$. where
$$I_1=\frac{1}{4\pi r^2}\int_0^{\pi}rd\theta \int_0^{2\pi}r\sin\theta d\phi[r \sin\theta\cos\phi]=\frac{r^3}{4\pi r^2}\int_0^{\pi}\sin^2\theta d\theta \int_0^{2\pi} \cos\phi d\phi=0$$
$$I_2=\frac{x_0}{4\pi r^2}\int_0^{\pi}rd\theta \int_0^{2\pi}r\sin\theta d\phi=x_0$$
Of course $u=0+x_0=\text{constant}$ satisfies the homogeneous PDE:
$$u_{rr} + \frac{2}{r} u_r = u_{xx} + u_{yy} + u_{zz}$$
