An area of the sphere in spherical coordinates is:
$$A=\int_0^{2\pi}\int_0^\pi d\Omega=\int_0^{2\pi}\int_0^\pi \sin \theta d\theta d\phi$$
Where $\Omega$ is a solid angle. Integral gives old good $4\pi$ as required.
Now, lets say that I want to weight every solid angle element with a unit vector:
$$\vec{r}=(\cos \theta \cos\phi, \cos \theta \sin \phi, \sin \theta)$$
This unit vector points to an area of a solid angle. For every vector $\vec{v}_0$ there would exist $-\vec{v}_0$ canceling each others contribution to the integral, therefore I would expect that due to spherical symmetry the following integral vanishes, however it does not. Result is:
$$\int_0^{2\pi}\int_0^\pi \vec{r} d\Omega = (0, 0, \pi^2) \neq \vec{0}$$
Where is my intuition failing?