The trick is in being able to draw polar plots roughly, from the spherical harmonics.
Consider the state, n=2, l=1, m=0
The polar part of the wavefunction is given by :
[tex]|Y_1^0|^2 = \frac{3}{4 \pi} cos^2 \theta[/tex]
Ignoring the constant term, the functional behavior is [itex]cos ^2 \theta[/itex], which we want to plot against [itex]\theta[/itex]
Let's draw the 2D version of this plot (or you can simply graph this using Mathematica or a calculator that can do polar plots). Draw the X (theta = 90, -90) and Z (theta = 0, 180) axes, and draw a large number of lines, all passing through the origin - like the spokes on a bicycle wheel. These lines represent the different values of theta. Now on each line, place a point at a distance (from the origin) given by [itex]cos^2 \theta[/itex]. Finally, join all these points, neighbor to neighbor. Remember, the X-axis represents [itex]\theta = 90[/itex] and the Z axis represents [itex]\theta = 0[/itex]
At [itex]\theta =90,~ cos^2 \theta = 0[/itex]. So, the points along the horizontal spokes are at the origin. At [itex]\theta[/itex] increases, or decreases, [itex]cos ^2\theta[/itex] increases till it reaches maxima at [itex]\theta = 0, 180[/itex]. So, for spokes above and below the X-axis, the points move farther and farther out, reaching a maximum at the +/-Z-axis. Join these points and you'll find it looks like a vertical dumbell oriented along the Z-axis. So, clearly this is the [itex]2p_z[/itex] orbital. While this is not all of it (you must now combine the radial part with the polar part), and the best way involves using some 3D plot software, I can't really do that here.
For the s-orbitals, the total wavefunction is something like [itex]|\psi _n(r, \theta \phi)|^2 ~=~A r^{2n}e^{-2r/na_0}[/itex]. Since these orbitals have no polar ([itex]\theta[/itex]) or azimuthal ([itex]\phi[/itex]) dependence, they are spherically symmetric, and all the "equipotential" surfaces are spheres.
If this was too confusing, I'll try and attach a picture, when I find a little more time.
PS : Not more clever...maybe a little more experienced, that's all !
