Why are points uniformly distributed on a sphere in 3D uniformly distributed in component coordinates? I've generated uniformly random points on a sphere (in 3D). As expected, all azimuthal angles are drawn with equal probability and it's less likely to draw points close to the poles:

However, when I histogram the Cartesian coordinates, I see that the draws are uniformly distributed in all dimensions: 

Why are the points uniformly distributed in the x, y, and z dimensions? I don't find that intuitive. Slices near the poles have less surface area than slices near the equator. Therefore, I expect more points on the equatorial slice, and for the points to cluster around the mean when I project them onto a single dimension (like the distribution of theta):

I generated uniformly random points on the surface of the sphere with 3 different methods because I was convinced I was doing something wrong, but they all gave the same results:
import numpy as np
from numpy.random import uniform

n = 500000

# python implementation of http://stackoverflow.com/a/6390021/4212158
# where theta0 => azimuth and theta1 => inclination
azimuth = uniform(0, 2*np.pi, n)
inclination = np.arccos(1 - 2*uniform(0, 1, n))
radius = 1.
x = radius * np.sin(inclination) * np.sin(azimuth)
y = radius * np.sin(inclination) * np.cos(azimuth)
z = radius * np.cos(inclination)

# rescaling draws from 3D gaussian method
# credit http://stackoverflow.com/a/33977530/4212158
gaussian_points = np.random.randn(3, n)
x, y, z = gaussian_points / np.linalg.norm(gaussian_points, axis=0)
azimuth = np.arctan2(y, x)
inclination = np.arccos(z)

# trig method 2
# credit http://stackoverflow.com/a/14805715/4212158
z = uniform(-1, 1, n)
azimuth = uniform(0, 2*np.pi, n)
x = np.sqrt(1 - z**2) * np.cos(t)
y = np.sqrt(1 - z**2) * np.sin(t)
inclination = np.arccos(z)

Python code for the plots:
import matplotlib.pyplot as plt
f, axarr = plt.subplots(3, sharex=True)
axarr[0].hist(x, bins=50, normed=True)
axarr[0].set_title('x')
axarr[1].hist(y, bins=50, normed=True)
axarr[1].set_title('y')
axarr[2].hist(z, bins=50, normed=True)
axarr[2].set_title('z')
plt.suptitle('Histogram of xyz coordinate draws')
plt.xlabel("Distance from origin")
plt.ylabel("Probability")
plt.show()


fig3 = plt.figure()
plt.scatter(azimuth, inclination, c='black', marker='.', alpha=.02)
plt.title("Azimuthal vs inclination angles of points uniformly distributed on surface of unit sphere")
plt.xlabel("Azimuthal angle (radians)")
plt.ylabel("Inclination angle (radians)")
plt.show()

fig = plt.plot()
plt.hist(inclination, bins=50, normed=True)
plt.title("Distribution of inclination angles")
plt.xlabel("Inclination angle (radians)")
plt.ylabel("Probabilities")
plt.show()

 A: Here's where your intuition fails you: slice the surface into rings of equal width along $x$ axes. Although the width along $x$ is the same $\Delta x$, the width of the surface of each band gradually increases as you move from $x=0$ to $x=1$, because the surface turns from horizontal to vertical. The angle change is just enough to compensate for shorter circumference of the rings closer to the poles. 
Here's my crappy diagram. You see how $dx$ is the same, but $AB$ is shorter than $CD$ (because it's at angle to $x$-axis) while $AA'$ is longer than $CC'$. In the end the surface of $AB$ band ends up being the same as $CD$ band. Magic!

Let me know if you want more detail or math. It's easy to show. Also exploring the difference between picking point on a sphere and inside a sphere is very interesting. Try getting the distributions of Cartesians for the latter case.
Here's MATLAB code and plots for these two cases and 3 dimensions.

rng(0,'twister')
rvals = 2*rand(1000,1)-1;
elevation = asin(rvals);

azimuth = 2*pi*rand(1000,1);
radii = 3*(rand(1000,1).^(1/3));

[x,y,z] = sph2cart(azimuth,elevation,3);
figure
subplot(2,2,1)
plot3(x,y,z,'.')
axis equal
title 'on sphere'

subplot(2,2,2)
hist(x)
title 'distribution of x on sphere'

[x,y,z] = sph2cart(azimuth,elevation,radii);
subplot(2,2,3)
plot3(x,y,z,'.')
axis equal
title 'inside sphere'

subplot(2,2,4)
hist(x)
title 'distribution of x inside sphere'

