Area of projection of ellipsoid onto the $xy$ plane

Question

Given the ellipsoid

$$ (\mathbf{r} - \mathbf{r_0} )^T Q (\mathbf{r} - \mathbf{r_0} ) = 1 $$

with $Q, \mathbf{r_0}$ known.

Question: Find the area of its projection onto the $xy$ plane.

If $a,b,c$ are the lengths of the three mutually perpendicular semi-axes of the ellipsoid, then

$ Q = R D R^T $

where $ D = \begin{bmatrix} \dfrac{1}{a^2} && 0 && 0 \\ 0 && \dfrac{1}{b^2} && 0 \\ 0 && 0 && \dfrac{1}{c^2} \end{bmatrix} $

And $R$ is a rotation matrix whose columns are the unit vectors along the directions of the three axes of the ellipsoid.

As an numerical example, let $a = 30, b = 15, c = 10 $, and let

$R = \dfrac{1}{9} \begin{bmatrix} 8 && 4 && 1 \\ -4 && 7 && 4 \\ 1 && -4 && 8 \end{bmatrix} $

and

$ \mathbf{r_0} = \mathbf{0} $

Find the numerical value of the area of the shadow of this ellipsoid onto the $xy$ plane.

My approach: is detailed in my answer below.

The numerical value of the area found there is $ A_{\text{projection}} = 1325.646 $

I would like to verify this value of the area. Your comments, and alternative solutions are highly appreciated.

Answer 1

There is a formula (see here) that I derived for the shadow of an ellipsoid in general orientation onto an arbitrary plane.

The area of this shadow is given by

$ A_{\text{shadow}} = \pi \dfrac{ \sqrt{ \mathbf{d_0}^T Q \mathbf{d_0} } }{ \sqrt{ | Q | } | \mathbf{n}^T \mathbf{d_0} | } $

where $d_0$ is the unit vector direction of the parallel light rays that are intercepted by the ellipsoid, and $\mathbf{n}$ is the normal vector to the shadow plane.

The area of the projection of the ellipsoid onto the $xy$ plane is a special case of the ellipsoid shadow. Here, we have

$ \mathbf{d_0} = \mathbf{n} = [ 0, 0, 1]^T $

Therefore, the area of the projection will be

$ A_{\text{projection}} = \pi \dfrac{ \sqrt{ Q_{33} }}{ \sqrt{ |Q| }} $

Since $ Q = R D R^T $ and $ R $ is a rotation matrix then,

$ | Q | = | D | $

Hence

$ A_{\text{projection}} = \pi a b c \sqrt{ Q_{33} } $

In our example,

$Q_{33} = \dfrac{1}{a^2} \bigg(\mathbf{u_1 u_1}^T\bigg)_{33} + \dfrac{1}{b^2} \bigg(\mathbf{u_2 u_2}^T\bigg)_{33} + \dfrac{1}{c^2} \bigg(\mathbf{u_3 u_3}^T\bigg)_{33}$

This is equal to

$ Q_{33} = \dfrac{1}{81} \left( \dfrac{1^2}{30^2} + \dfrac{(-4)^2}{15^2 } +\dfrac{8^2}{10^2} \right)= \dfrac{641}{72900} $

Therefore, the area of the projection according to the above formula is,

$ \boxed{A_{\text{projection}} = \pi (30)(15)(10) \sqrt{ \dfrac{641}{72900} } \approx 1325.646 }$