What is the average cross section area $A_n$ of a cuboid normal to a vector $\vec{n}$ and its height $H_n$ along the normal?

Question

It is a followup to my previous question of an average chord length in a rectangle: What’s the average length of chords at a given angle in a rectangle?

Now, I am faced with a 3D body, I need some help in obtaining the average cross-section area $A_n$ of a cuboid normal to a vector $\vec{n}$ and its height $H_n$ along the normal? For simplicity, you can assume $\vec{n}$ was obtained by rotation of $\alpha$ around the current z-axis, then by $\beta$ around the current y-axis, giving $\vec{n}=\vec{x} \cos\alpha\cos\beta - \vec{y} \sin\alpha + \vec{z} \cos\alpha \sin\beta$

Answer 1

I'll assume that $\vec{n}$ is a unit vector throughout: otherwise, normalise it first.

Echoing the answer from your previous question, the volume is the height (along your $\vec{n}$) multiplied by the average cross-sectional area. Thus, the average cross sectional area is exactly the volume $HLW$ divided by the height along $\vec{n}$.

We therefore only need to calculate the height of the cube along $\vec{n}$.

For this purpose, we will need an orthonormal basis with $\vec{n}$ as its first element. To achieve this, we'll simply take the standard basis and rotate it as you say at the end of your question. The change of basis matrix from the standard basis to this new one is then $$\left(\begin{array}{ccc}\cos\beta & 0 & \sin\beta \\ 0 & 1 & 0 \\ -\sin\beta & 0 & \cos\beta\end{array} \right)\left(\begin{array}{ccc}\cos\alpha & -\sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1\end{array}\right) = \left(\begin{array}{ccc}\cos\alpha\cos\beta & -\sin\alpha\cos\beta & \sin\beta \\ \sin\alpha & \cos\alpha & 0 \\-\cos\alpha\sin\beta & \sin\alpha\sin\beta & \cos\beta\end{array}\right).$$

Now, in our original coordinates, the vector between the most down-$\vec{n}$ and up-$\vec{n}$ corners of the cube is (WLOG - else rotate things around such that all components of $\vec{n}$ are positive) $$\left(\begin{array}{c}L \\ H \\ W\end{array}\right).$$ The height is then exactly the $\vec{n}$ coordinate of the image of this under the above change of basis. That is, it is $$L\cos\alpha\cos\beta - H\sin\alpha\cos\beta + W\sin\beta.$$

Thus, the average cross-sectional-area is precisely $$\dfrac{HWL}{L\cos\alpha\cos\beta - H\sin\alpha\cos\beta + W\sin\beta}.$$

Answer 2

Assuming the height along $\vec{n}$ follow equation of ellipsoid with semi-axes $(L,H,W)$. The height along $\vec{n}$ can be written as,

\begin{align} \sqrt{L^2 \cos^2\alpha \cos^2\beta+H^2 \sin^2\alpha+W^2 \cos^2\alpha \sin^2\beta} \end{align}

Hence, the average cross-sectional-area is

\begin{align} \frac{H W L}{\sqrt{L^2 \cos^2\alpha \cos^2\beta+H^2 \sin^2\alpha+W^2 \cos^2\alpha \sin^2\beta}} \end{align}