The spherical mean of a function $h : \mathbb R^2 \to \mathbb R$ is given by $$ \frac{1}{2\pi r} \int_0^{2\pi} h(x + r \cos(\theta), y + r \sin(\theta)) d \theta $$ Now I want to compute the spherical mean of the function $h(x,y) = x$, it is $$ \frac{1}{2\pi r} \int_0^{2\pi} (x + r\cos(\theta)) d \theta = \frac{1}{2\pi r}\int_0^{2\pi} x d\theta + \frac{1}{2\pi r} \int_0^{2\pi} r \cos(\theta) d\theta = \frac{x}{r} + 0 = x/r. $$ But I do not trust this result, first if I image this function (it describes a plane rotated by $45$ degrees to the $x$-plane) the average around some point should equal the $x$-coordinate at that point.
Further this function is certainly harmonic, i.e. it satisfies $\Delta h = 0$, and for such functions the spherical mean at some point equals the value of the function at that point, which is not satisfied here. So whats wrong?
There is a mistake. The length element ($ds(p)$) of the circle of radius $r$ is not (in polar coordinates) $d\theta$ but $r\,d\theta$. You have divided twice by $r$, but it should be only once. The mean is
$$\frac{1}{2\pi}\int_0^{2\pi} h(x+r\cos\theta, y+r\sin\theta)\,d\theta,$$
and for $h(x,y) = x$, you get the mean $x$, as it should be.