A closed form expression for shadow area of an ellipsoid subject to uniform direction light

The shadow of the ellipsoid on the given plane is generated by the intersection of the plane and the elliptical cylinder that is tangent to the ellipsoid, and which has its axis parallel to the fixed direction of light rays $\mathbf{d_0}$ .

Since the normal vector to the surface of the ellipsoid is given by $2 Q_e ( \mathbf{r} - \mathbf{r_0} )$ , it follows that tangency points on the ellipsoid are characterized by the vector equation,

$\mathbf{d_0}^T Q_e (\mathbf{r} - \mathbf{r_0} ) = 0$

This is an equation of a plane passing through the center of the ellipsoid $\mathbf{r_0}$ and having a normal vector of $Q_e \mathbf{d_0}$ . We will now take a different approach to finding the area $A_0$ of the ellipse of intersection between this plane and the ellipsoid. Consider a family of cutting planes all having the unit normal $\mathbf{n_1} = \dfrac{ Q_e \mathbf{d_0} }{\sqrt{ \mathbf{d_0}^T Q_e^2 \mathbf{d_0} } }$ and passing through the point $\mathbf{p_1}(t) = \mathbf{r_0} + t \mathbf{n_1}$ .

If $V(t)$ is the volume of the portion of the ellipsoid that lies on the opposite side of the plane as the side pointed to by the normal vector $\mathbf{n_1}$ , then the area of the cut $A(t)$ is equal to the derivative of $V(t)$ with respect to $t$ . Now it is rather simple to find $V(t)$ . To this end, we scale down the ellipsoid to a unit sphere, as follows:

Let the symmetric matrix $Q_e$ be decomposed as follows:

$Q_e = R D R^T = R \sqrt{D} \sqrt{D} R^T$

where $D$ is a diagonal matrix of $Q_e$ eigenvalues, and $R$ is a rotation matrix. Now define $\mathbf{r'} = \sqrt{D} R^T (\mathbf{r} - \mathbf{r_0} )$ , then it follows that $\mathbf{r'}^T \mathbf{r'} = 1$ , that is, $\mathbf{r'}$ lies on the unit sphere with center at the origin. The original cutting plane is the plane, $\mathbf{n_1}^T (\mathbf{r} - (\mathbf{r_0} + t \mathbf{n_1}) ) = 0$ . The transformation of this plane into the $\mathbf{r'}$ space is obtained by substituting,

$\mathbf{r} = \mathbf{r_0} + R \sqrt{D^{-1}} \mathbf{r'}$

into the equation of the plane; this results in:

$\mathbf{n_1}^T (R \sqrt{D^{-1}} \mathbf{r'} - t \mathbf{n_1}) = 0$

The distance between this plane (now in the $\mathbf{r'}$ space) , and the origin is $a = \dfrac{t}{ \sqrt{ \mathbf{n_1}^T R D^{-1} R^T \mathbf{n_1}} }$ . But $\mathbf{n_1} = \dfrac{ Q_e \mathbf{d_0} }{\sqrt{ \mathbf{d_0}^T Q_e^2 \mathbf{d_0} } }$ , it follows that the distance $a = t \dfrac{\sqrt{ \mathbf{d_0}^T Q_e^2 \mathbf{d_0} } }{ \sqrt{ \mathbf{d_0}^T Q_e \mathbf{d_0} } } = t a_0$ .

Now the volume of the portion of the unit sphere lying on the opposite side of the side pointed to by the normal $\mathbf{n'} = \sqrt{D^{-1}} R^T \mathbf{n_1}$ is given by (this is very simple to prove):

$V_{1}(t) = \pi ( \dfrac{2}{3} + a - \dfrac{1}{3} a^3 ) = \pi ( \dfrac{2}{3} + a_0 t - \dfrac{1}{3} a_0^3 t^3 )$

The corresponding volume of the ellipsoid is simply

$V(t) = | \sqrt{ D^{-1} } | V_1(t) = \sqrt{ | D^{-1} | } V_1(t) = \dfrac{V_1(t)}{ \sqrt{ | D | }} = \dfrac{V_1(t)}{\sqrt{ | Q_e | } }$

Note that the cutting plane we're interested in is at $t = 0$ , therefore, we differentatiate $V(t)$ with respect to $t$ and evaluate the derivative at $t = 0$ , therefore,

$A_0 = \dfrac{\pi a_0}{\sqrt{ | Q_e | }} = \dfrac{\pi}{\sqrt{ | Q_e | } } \dfrac{ \sqrt{ \mathbf{d_0}^T Q_e^2 \mathbf{d_0} } }{ \sqrt{ \mathbf{d_0}^T Q_e \mathbf{d_0} } }$

We're half way to find the desired expression. Note that the normal to the cut plane is along $\mathbf{n_1}$ , which is parallel to $Q_e \mathbf{d_0}$ while the axis of the cylinder which is orthogonal to its cross section is along $\mathbf{d_0}$ , thus if $\theta_1$ is the angle between these two vectors then the area of the cross section of the elliptical cylinder is given by

$A_c = A_0 | \cos \theta_1 |$

Now, using the inner product, we know that $\cos \theta_1 = \dfrac{ \mathbf{d_0}^T Q_e \mathbf{d_0} }{\sqrt{ \mathbf{d_0}^T Q_e^2 \mathbf{d_0} } }$ (remember that $\mathbf{d_0}$ is a unit vector ).

Finally, the shadow area is related to the cross sectional area of the shadow cylinder by

$A_{\text{shadow}} = \dfrac{ A_c }{| \cos \theta_2 |}$

where $\theta_2$ is the angle between the axis of the cylinder ( $\mathbf{d_0}$ ) and the normal to the projection plane ( the vector $\mathbf{n}$ ) . So that $\cos \theta_2 = \mathbf{n}^T \mathbf{d_0}$ .

Putting it all together, we obtain

$A_{\text{shadow}} = \dfrac{\pi}{\sqrt{ | Q_e | }} \dfrac{\sqrt{ \mathbf{d_0}^T Q_e^2 \mathbf{d_0} } }{\sqrt{ \mathbf{d_0}^T Q_e \mathbf{d_0} } } \dfrac{ \mathbf{d_0}^T Q_e \mathbf{d_0} }{\sqrt{ \mathbf{d_0}^T Q_e^2 \mathbf{d_0} }} \dfrac{1}{| \mathbf{n}^T \mathbf{d_0} | }$

Simplifying, we obtain,

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

Hence, the correct choice is B.