The average area of the shadow of a square This question comes from my attempt at solving the problem of the average area of a cube in an unconventional way. As the shadow of the cube is comprised of the shadows of 3 of its square faces, I am attempting to calculate the average area of each of those squares' shadows, multiply it by three, and arrive at the answer of $\frac{3s^2}{2}$. That means that the average area of the shadow of a square should be $\frac{s^2}{2}$, which makes intuitive sense, but I want to prove it mathematically, which I am having difficulty doing. Here's my work so far:
Consider a square $\mathrm{S}$ with side lengths $s$ that is lifted up in both the $\mathrm{x}$ and $\mathrm{y}$ directions by $\theta_1$ and $\theta_2$ degrees respectfully. Then, for these $\theta$ values, the area of the square becomes $s^2\mathrm{cos}\theta_1\mathrm{cos}\theta_2$, as can be determined by trigonometry (a right triangle with a hypotenuse of $s$, angle $\theta$, and side $x$ adjacent to the angle gives $x=s\mathrm{cos}\theta$). The median value of $\mathrm{cos}\theta$ on the interval $[0, \pi/2]$ is $\frac{\sqrt{2}}{2}$ (appearing at $\frac{\pi}{4}$).
Here's the part I don't fully understand:
Let $\mathrm{A(S_{(\theta_a, \theta_b)})}$ be a function of a square $\mathrm{S}$ that has been lifted by $\theta_a$ and $\theta_b$ on the $\mathrm{x}$-axis and $\mathrm{y}$-axis respectively. To find the average area of this square, we can do the following: $$\frac{\iint_R \mathrm{A(S_{(\theta_a, \theta_b)})}dA}{R_{area}}, \{R\in\mathbb{R}^2 \vert R = [0,\frac{\pi}{2}] \times [\frac{\pi}{2}]\} \\ = \frac{s^2\int_0^{\pi/2}\int_0^{\pi/2}\mathrm{cos}\theta_1\mathrm{cos}\theta_2d\theta_1d\theta_2}{\frac{\pi^2}{4}} \\ = \frac{4s^2}{\pi^2}$$
This does not yield the result I'm looking for, so I'm wondering where I messed in my logical deduction. Can I just take the median value of $\mathrm{cos}x$ over $[0,\frac{\pi}{2}]$? If so, that removes the need for the integral, but it seems to simple, and doesn't explain why the two contradict each other. Thanks in advance.
 A: There are several problems with this, perhaps related to each other.

*

*You average the orientations over the planar cross-sectional area. You should be doing so over the surface area of a sphere to which the square is presumed tangent. This surface area over one octane (both $\theta$ values between $0$ and $\pi/2$) is $\pi/2$ instead of $\pi/4$.


*Also with the spherical averaging, you need to observe that the probability distribution is biased in favor of the "equator" because a strip there covers more incremental area than an equally wide strip near the "pole". The strip near the equator is longer. The extra factor that accounts for this is the sine of the latitude, with the latitude being measured from the pole.


*Finally, the projection factor is dependent on just the latitude, specifically the cosine of this quantity. With the projection lines coming from the pole to the center of the sphere and parallel lines to that, rotating the square longitudinally dies not really affects its area.
So, rendering $\theta_1$ as latitude measured from the pole and $\theta_2$ as longitude, your integral should properly read
$\dfrac{s^2\int_0^{\pi/2}\int_0^{\pi/2}(\cos\theta_1\sin\theta_1)d\theta_1d\theta_2}{\pi/2}$
This should give $(1/2)s^2$.
