Why is the dz2 orbital so different from the rest?
The wikipedia is helpful in explaining why radial variations should arise in the density of non-s orbitals:


  The non radial-symmetry properties of non-s orbitals are necessary to localize a particle with angular momentum and a wave nature in an orbital where it must tend to stay away from the central attraction force, since any particle localized at the point of central attraction could have no angular momentum. 




What's unique about the $d_{z^2}$ orbital (see the table above, from the wikipedia) compared to the other $l=2$ angular momentum wavefunctions is that the z-component is zero ($m=0$). This further constrains the geometry of the wavefunction.

The functions describing the angular dependence of the hydrogenic wavefunctions are Legendre polynomials $Y_{lm}(\theta,\phi)$, solutions of Legendre's differential equation. In the case of d-orbitals, they satisfy

$$ \hat{L}^2Y_{lm}(\theta,\phi)=\hbar^2l(l+1)Y_{lm}(\theta,\phi)$$

with $l=2$, where $ \hat{L}$ is the angular momentum operator. Since the z-component of the angular momentum is also quantized, the following eigenequation also holds:

$$ \hat{L}_zY_{lm}(\theta,\phi)=\hbar mY_{lm}(\theta,\phi)$$

with $m=0$ in the case of the $d_{z^2}$ orbital, and this last equation leads to the following condition:

$$ \frac{\partial\psi}{\partial y^2} = \frac{\partial\psi}{\partial x^2}$$

which implies that the solutions must be cylindrically symmetric about z. However, the condition $l\neq 0$ implies that the solution is not spherically symmetric. The result is the unexpected shape of the $d_{z^2}$ orbital.