How to find average shadow of a unit cube with the formula of area of orthogonal projection of cube? I attempted to find average shadow of cube using the formula of
Orthogonal Projection Area of a 3-d Cube
$$\sin \alpha + \cos \alpha(\sin\beta + \cos\beta)$$
However I was not able to reach the answer which is 1.5. Where did I go wrong in my attempt ?
My attempt:
Using the formula for orthogonal projection. I tried the following integration to find the average.
$$\frac{\int_0^{\pi/2}\int_0^{\pi/2}\sin \alpha + \cos \alpha(\sin\beta + \cos\beta) d\alpha d\beta}{\pi^2/4}$$
This however did not yield the correct answer. Why does this not work? What is the correct method to this. Thanks  in advance
$\alpha$ and $\beta$
This contains the image of what the angles alpha and beta represent
 A: Assume the cube is initially in upright position.  Then you apply two rotations to it, the first one, about its own $y$ axis, and the second one about the resulting cube's own $z$ axis.  The corresponding rotation matrix due to these two rotations is given by $R_1 R_2 $ (because we're using relative rotations, i.e. rotations not about world axes, but local (cube) axes)
Now,
$R_1 = \begin{bmatrix} \cos \theta_1 && 0 && \sin \theta_1 \\ 0 && 1 && 0 \\
-\sin \theta_1 && 0 && \cos \theta_1 \end{bmatrix} $
and
$R_2 = \begin{bmatrix} \cos \theta_2 && - \sin \theta_2 && 0 \\ \sin \theta_2 && \cos \theta_2 && 0 \\ 0 && 0 && 1 \end{bmatrix} $
so that $R_1 R_2 = \begin{bmatrix} \cos \theta_1 \cos \theta_2 && - \cos \theta_1 \sin \theta_2 && \sin \theta_1 \\
\sin \theta_2 && \cos \theta_2 && 0 \\ - \sin \theta_1 \cos \theta_2 && \sin \theta_1 \sin \theta_2 && \cos \theta_1 \end{bmatrix} $
The columns of $R_1 R_2 $ are the normal vectors to three non-parallel faces, and the area of the projection is the sum the areas of each face which is $1$ times the absolute cosine of the angle between each of these column vectors and the vector $k = [0, 0, 1]^T $ (Assuming projection is onto the $xy$ plane)
Hence
$A(\theta_1, \theta_2) = | \sin \theta_1 \cos \theta_2 | + | \sin \theta_1 \sin \theta_2 | + | \cos \theta_1 |$
Note that $\theta_1 \in [0, \pi] $ and $\theta_2 \in [0, 2 \pi] $
Hence, the average area is
$\overline{A} =   \displaystyle \dfrac{\displaystyle \int_0^\pi \int_0^{2 \pi} \sin \theta_1 \left(\sin \theta_1 ( | \cos \theta_2 | + | \sin \theta_2 | ) + | \cos \theta_1 | \right) \text{d} \theta_2 \text{d} \theta_1 }{ \displaystyle \int_0^\pi \int_0^{2\pi} \sin \theta_1\text{d} \theta_2 \text{d} \theta_1 } $
Note that $\displaystyle \int_0^{2\pi} (( | \cos \theta_2 | + | \sin \theta_2 | ) \text{d} \theta_2 = 4(2) = 8 $
Thus the above integral becomes
$\overline{A} = \dfrac{1}{4 \pi} \displaystyle \int_0^\pi 8 \sin^2 \theta_1 + (2 \pi) \sin \theta_1| \cos \theta_1 |\text{d} \theta_1 $
And this evaluates to
$\overline{A} = \dfrac{4 \pi + (2 \pi) }{ 4 \pi } = 1.5 $
