Shadow of a disc inside a cube

Let the center of the cube be $(0, 0, 0)$ and the light source be $A(3, 3, 3)$ . Let $B(p, q, 0)$ be a point on the edge of the disc that is projected to $C(-3, y, z)$ on the far wall of the cube.

Since $B(p, q, 0)$ is on the disc with a radius of $3$ , $p^2 + q^2 = 9$ , or $q = \sqrt{9 - p^2}$ .

The vector $AB$ has an equation of $(x, y, z) = (p - 3, q - 3, 0 - 3)t + (3, 3, 3)$ . $C(-3, y, z)$ is also on that vector, so from the $x$ -coordinate we have $-3 = (p - 3)t + 3$ , which solves to $t = -\cfrac{-6}{p - 3}$ . Therefore, $y = (q - 3)t + 3 = \cfrac{-6(q - 3)}{p - 3} + 3$ and $z = (0 - 3)t + 3 = \cfrac{18}{p - 3} + 3$ .

Substituting $q = \sqrt{9 - p^2}$ into $y = \cfrac{-6(q - 3)}{p - 3} + 3$ gives $y = \cfrac{-6(\sqrt{9 - p^2} - 3)}{p - 3} + 3$ , and this with $z = \cfrac{18}{p - 3} + 3$ after eliminating $p$ and simplifying gives $z = y - 2 \sqrt{3} \sqrt{3 - y} - 6$ .

The area of the shadow on one of the far walls is then $A_{\text{wall}} = \displaystyle \int_{-3}^{3} (y - 2 \sqrt{3} \sqrt{3 - y} - 6 + 3) dy = 24\sqrt{2} - 18$ .

The area of the shadow on the floor is one fourth of a circle with a radius of $6$ , or $A_{\text{floor}} = \frac{1}{4}\pi 6^2 = 9 \pi$ .

The total area is therefore $A_{\text{total}} = A_{\text{floor}} + 2A_{\text{wall}} = 9\pi + 2(24\sqrt{2} - 18) = 9\pi + 48\sqrt{2} - 36 \approx \boxed{60.156}$ .

Let the cube lie in the first octant with one corner at the origin, and the opposite corner at $(6, 6, 6)$ , and let the light source be at $\mathbf{A} = (0, 0, 6)$ . The disc is centered at $\mathbf{H} = (3, 3, 3)$ , and its circular edge has the parametric equation $\mathbf{q}(t) = \mathbf{H} + \cos t \mathbf{v_1} + \sin t \mathbf{v_2}$ , where $\mathbf{v_1} = (3, 0, 0)$ and $\mathbf{v_2} = (0, 3, 0)$ . The rays from the source to the edge of the disc are given by

$\mathbf{r} = \mathbf{A} + s \mathbf{M v}$

where $s \ge 0$ is a scalar, and $\mathbf{v} = [ \cos t , \sin t , 1 ]^T$ and

$\mathbf{M} = [ \mathbf{v_1} , \mathbf{v_2} , \mathbf{H - A} ] = \begin{bmatrix} 3 && 0 && 3 \\ 0 && 3 && 3 \\ 0 && 0 && -3 \end{bmatrix}$

Noting that $\mathbf{v}^T \mathbf{Q_0} \mathbf{v} = 0$ , where $\mathbf{Q_0} = \begin{bmatrix} 1 && 0 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && -1 \end{bmatrix}$

It follows that, the rays lie on a cone whose equation is

$(\mathbf{r} - \mathbf{A} )^T \mathbf{M}^{-T} \mathbf{ Q_0 } \mathbf{M}^{-1} (\mathbf{r} - \mathbf{A} ) = 0$

The inverse of matrix $\mathbf{M}$ is easily computable to be,

$\mathbf{M}^{-1} = \dfrac{1}{3} \begin{bmatrix} 1 && 0 && 1 \\ 0 && 1 && 1 \\ 0 && 0 && -1 \end{bmatrix}$

Hence,

$\mathbf{Q_c} = \mathbf{M}^{-T} \mathbf{ Q_0 } \mathbf{M}^{-1} = \dfrac{1}{9} \begin{bmatrix} 1 && 0 && 1 \\ 0 && 1 && 1 \\ 1 && 1 && 1 \end{bmatrix}$

so that the equation of the cone is

$(\mathbf{r} - \mathbf{A})^T \mathbf{Q_c} (\mathbf{r} - \mathbf{A}) = 0$

intersecting this cone with the plane $z = 0$ , gives

$\mathbf{r} - \mathbf{A} = \begin{bmatrix} x \\ y \\ -6 \end{bmatrix}$

so that the equation of the intersection is,

$x^2 + y^2 + 36 - 12 x - 12 y = 0$

by completing the squares for $x$ and $y$ , this becomes,

$(x - 6)^2 + (y - 6)^2 = 36$

which is the equation of a circle centerd at $(6, 6, 0)$ with a radius $6$ , hence the area bounded by this circle and the edges of the cube is $A_1 = \frac{1}{4} \pi (36) = 9 \pi$

Next, intersecting the cone with the plane $y = 6$

$\mathbf{r} - \mathbf{A} = \begin{bmatrix} x \\ 6 \\ z -6 \end{bmatrix}$

so that the equation of the intersection is,

$x^2 + 36 + (z - 6)^2 + 2 x (z - 6) + 12 (z - 6) = 0$

which is a quadratic in $(z - 6)$ , its solution (using the quadratic formula) is

$(z - 6) = \frac{1}{2} [ - (2x + 12) \pm \sqrt{ (2x + 12)^2 - 4 (x^2 + 36) }]$

$= \frac{1}{2} [ - (2x + 12) \pm \sqrt{48 x } ]$

$= - (x + 6) \pm 2 \sqrt{3 x}$

so that

$z = - x \pm 2 \sqrt{3x}$

since we want $z \ge 0$ , then $z = - x + 2 \sqrt{3 x }$

now the area is $A_2 = \displaystyle \int_ {x = 0}^{x = 6} (- x + 2 \sqrt{3x})dx = -18 + 24 \sqrt{2}$

Since this is the same area of the shadow on the plane $x = 6$ , then the total shadow area is $A_{\text{Shadow}} = A_1 + 2 A_2 = 9 \pi - 36 + 48 \sqrt{2} \approx \boxed{60.156}$