Spinning surfaces and the vertical lines that test them

In a more advanced course, you’ve been asked to draw a 3D sketch of the function $z = |x| + |y|.$ As you’re about to do this, you are struck by another bout of dizziness, and your resulting graph is rotated about the origin in a random direction in 3D space. In other words, the central axis of your graph’s “funnel” is equally likely to point from the origin to any point on the surface of the unit sphere.

What is the probability that the resulting graph you produce is in fact a function (i.e., $z$ is a function of $x$ and $y$)?

At first glance, this looks like a harder problem. There is after all one more direction to dizzily rotate your surfaces. However, first we note that the orthonormal rotation that transforms the point unit $z$-vector, $e_3 = (0,0,1)^T$ to a random point on the unit sphere $$\hat{u} = ( \cos \theta \sin \vartheta, \sin \theta \sin \vartheta, \cos \vartheta)$$ is composed of a rotation about the $y$-axis through an angle $\vartheta \in ( 0^\circ, 180^\circ),$ followed by a rotation about the $z$-axis through an angle of $\theta \in (0^\circ, 360^\circ).$ Now if we are testing from a vertical line test perspective using a vertical line parallel to the $z$-axis, then any rotation about the $z$-axis will not affect its success at passing this test. So we only are really worried about the rotation about the $y$-axis through an angle of $\vartheta.$ This case then devolves into the two-dimensional case we saw before, though with only half of range. That is, the resulting parametric surface will pass the vertical line test whenever $\vartheta \in [0^\circ, 45^\circ) \cup (135^\circ, 180^\circ].$

The only thing that we have to be careful here of is assessing the probability. Here we want to measure the surface of the sphere that satisfies $$\Omega = \{ \vartheta \in [0^\circ, 45^\circ) \cup (135^\circ, 180^\circ], \theta \in [0^\circ, 360^\circ) \}.$$ We can calculate the surface area of the region $\Omega$ by converting to radians and then using the surface area of the shape bounded by the curve $r = \sin \vartheta$ for $\vartheta \in [0, \frac{\pi}{4}) \cup (\frac{3\pi}{4}, \pi],$ that is $$S = S(\Omega) = 2 \int_0^{\pi / 4} \pi \sin^2 \vartheta \,d\vartheta.$$ Therefore, since the total surface area of the unit sphere is $4\pi,$ the probability of the rotated parametric surface passing the vertical line test is \begin{align*}q = \frac{S}{4\pi} &= \frac{1}{2} \int_0^{\pi/4} \sin^2 \vartheta \,d\vartheta \\ &= \frac{1}{2} \int_0^{\pi/4} \frac{1 - \cos 2\vartheta}{2} \,d\vartheta \\ &= \left[\frac{\vartheta}{4} - \frac{\sin 2\vartheta}{8} \right]^{\vartheta = \pi/4}_{\vartheta=0} \\ &= \frac{\pi}{16} - \frac{1}{8} = \frac{\pi-2}{16} \approx 7.13495\dots \%.\end{align*}