Calculating volume of spherical wedge from parallelepiped corner

Question

I am interested in calculating the volume of the intersection of a sphere of radius 1/2 with the corner of a parallelepiped where the angles between each edge is $\pi/3$ and has unit edge length; we have the center of the sphere placed on the corner vertex of the parallelepiped.

I tried approaching the problem with a triple integral in spherical coordinates but was not able to obtain the desired answer. Any ideas for a solution or a technique that generalizes to calculating the volume of the intersection of a sphere with a paralleletope in higher dimensions?

Answer 1

Apparently what you have is a corner of a regular tetrahedron, and you want to know what percentage of the volume (or surface area) of a little ball around one vertex is enclosed by the tetrahedron.

The first thing is that we need to know the "dihedral angle" between two faces of the tetrahedron. This is not hard; we can place one at every other vertex of a cube. That is, let the planes containing the four faces have normal vectors $$ (1,1,1), (-1,1,1), (1,-1,1), (1,1,-1). $$ The angle between the first two vectors is $$ \arccos \frac{1}{3} $$ or about $70^\circ.$

Next, http://en.wikipedia.org/wiki/Spherical_excess#Area_and_spherical_excess for a triangle on the unit sphere with three such angles, the area is $$ 3 \arccos \frac{1}{3} - \pi. $$ The total surface area is $4 \pi,$ so the fraction of the surface area is $$ (3 \arccos \frac{1}{3} - \pi) / (4 \pi) \approx 0.0439. $$ This is about one part in 23, just a little smaller than the size of a face of a spherical icosahedron.

Your sphere of radius 1/2 has volume $\frac{\pi}{6}.$ You want about 0.439 times that, or $$ \frac{3 \arccos \frac{1}{3} - \pi}{4 \pi} \frac{\pi}{6} = \frac{3 \arccos \frac{1}{3} - \pi}{24} $$

From $\cos 3 \theta = 4 \cos^3 \theta - 3 \cos \theta,$ the above is $$ \frac{ \arccos \frac{23}{27} }{24} $$

Answer 2

Let $\vec{a}, \vec{b}, \vec{c}$ be any 3 unit vectors. Let $T$ be the region span by the 3 rays starting at origin in the direction of $\vec{a}, \vec{b}$ and $\vec{c}$, i.e. $$T = \big\{\; \alpha \vec{a} + \beta \vec{b} + \gamma \vec{c} : \alpha,\beta,\gamma \ge 0\;\big\}$$ Let $B_r$ be the ball centered at origin with radius $r$.

The volume of the intersection between $B$ and $T$ is given by

$$ \text{Volume}( T \cap B ) = \frac13\Omega r^3$$

where $\Omega$ is the solid angle of $T$ as seen from the origin. $\Omega$ can be computed using following formula by Oosterom and Strackee:

$$\tan\left(\frac12\Omega\right) = \frac{\left|\vec{a}\cdot \left(\vec{b} \times \vec{c}\right)\right|}{ 1 + \vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}}$$

Please note that we have an algebraic identity $$\left|\vec{a}\cdot \left(\vec{b} \times \vec{c}\right)\right|^2 = \left|\begin{matrix} \vec{a}\cdot\vec{a} & \vec{a}\cdot\vec{b} & \vec{a}\cdot\vec{c}\\ \vec{b}\cdot\vec{a} & \vec{b}\cdot\vec{b} & \vec{b}\cdot\vec{c}\\ \vec{c}\cdot\vec{a} & \vec{c}\cdot\vec{b} & \vec{c}\cdot\vec{c}\\ \end{matrix}\right| $$ If the angles among the 3 rays are equal, i.e. $\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{c} = \vec{b}\cdot\vec{c} = \cos\theta$, we will have $$\left|\vec{a}\cdot \left(\vec{b} \times \vec{c}\right)\right|^2 = \left|\begin{matrix} 1 & \cos\theta & \cos\theta\\ \cos\theta & 1 & \cos\theta\\ \cos\theta & \cos\theta & 1\\ \end{matrix}\right| = 2\cos\theta^3 - 3\cos\theta^2 + 1 $$ This leads to $$ \tan\left(\frac12\Omega\right) = \frac{\sqrt{2\cos\theta^3 - 3\cos\theta^2 + 1}}{1 + 3\cos\theta}$$ Apply this to our case, we have $r = \frac12$, $\cos\theta = \cos\frac{\pi}{3} = \frac12$ and hence $\displaystyle\;\tan\left(\frac12\Omega\right) = \frac{\sqrt{2}}{5}\;$.
As a result,

$$\text{Volume} = \frac13\left(2\tan^{-1}\frac{\sqrt{2}}{5}\right) \left(\frac12\right)^3 = \frac{1}{12} \tan^{-1}\frac{\sqrt{2}}{5} \approx 0.022970233268022 $$