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Given a sphere of radius $r$, and a cube of unit length with the same center as the sphere, what is the surface area of the sphere that is inside the cube, when $\frac{\sqrt{2}}{2} <r < \frac{\sqrt{3}}{2}$?

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  • $\begingroup$ This question is a bit confusing, at least for me. $\endgroup$ – Ivan Lerner Aug 20 '13 at 20:39
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    $\begingroup$ What is meant by "a cube of unit length"? $\endgroup$ – AD. Aug 20 '13 at 20:52
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The derivation is messy, I hope I didn't make any mistake.

Let us view the unit sphere $S^2$ as a subset of $\mathbb{R}^3$.
For any geometric shape $X \subset S^2$, let $\Omega(X)$ be the corresponding solid angle/surface area.
For any $\rho \in (0,1)$ and unit vectors $\hat{n} \in S^2$, let $H_{\hat{n}}(\rho)$ be the half space $\{\; \vec{x} \in \mathbb{R}^3 : \vec{x} \cdot \hat{n} \ge \rho\;\}$.

It is well known the intersection of $S^2$ with such a half space $H_\hat{n}(\rho)$ is a spherical cap: $$\text{Cap}_{\hat{n}}(\rho) \stackrel{def}{=} S^2 \cap H_\hat{n}(\rho)$$ and the corresponding solid angle/surface area is independent of directions of $\hat{n}$:

$$\Omega(\text{Cap}_{\hat{n}}(\rho) ) = 2\pi (1-\rho)\tag{*1}$$

Let $\hat{x}$ and $\hat{z}$ be any two unit vectors on $S^2$ perpendicular to each other.
When $\rho < \frac{1}{\sqrt{2}}$, $\;\text{Cap}_{\hat{x}}(\rho)\;$ and $\;\text{Cap}_{\hat{z}}(\rho)\;$ intersect in a "lune" shaped object:

$$\text{Lune}_{\hat{x}\hat{z}}(\rho) \stackrel{def}{=} \text{Cap}_{\hat{x}}(\rho) \cap \text{Cap}_{\hat{z}}(\rho) = S^2 \cap H_{\hat{x}}(\rho) \cap H_{\hat{z}}(\rho)$$

Once again, the solid angle/surface area of these "lune" shaped objects is independent of directions of $\hat{x}$ and $\hat{z}$. For further discussion,let us assume $\hat{x}, \hat{y}$ are the unit vectors parallel to $x$ and $z$-axis.

If we project this lune to the plane $z = \rho$, we will obtain a figure composed of a circular arc and circular chord (the blue arc/chords) as shown in figure at end. Let $2\theta$ be the angle span by the arc/chord in the plane $z = \rho$, one find:

$$\cos\theta = \frac{\rho}{\sqrt{1-\rho^2}}$$

To compute the area of these "lune", we recall a well known result in spherical geometry:

Given any geodesic polygon $X$ on $S^2$. The solid angle/surface area enclosed by $X$ is equal to the difference of $2\pi$ and the sum of all exterior angles of the geodesic polygon.

The boundary of the "lune" is not a geodesic polygon on $S^2$. However, it doesn't stop us from approximating it with one. If one do the same thing to a spherical cap and compare it with the formula we have in $(*1)$. One can see that the "sum of exterior angles" along any one of the two "non-geodesic" edge between $P$ and $Q$ is simply $2\theta \rho$ $\color{blue}{^{[1]}}$.

The leaves us with the exterior angle at $P$ and $Q$. If one work out the tangent vectors of the two edges at $P$ in $\mathbb{R}^3$, they are just $( -\sin\theta, \cos\theta, 0)$ and $( 0, \cos\theta, -\sin\theta)$. This means if $\psi$ is the angle between these two tangent vectors, then

$$\cos\psi = \cos\theta^2 = \frac{\rho^2}{1-\rho^2}$$

and the change of exterior angle in $P$ or $Q$ is $\pi - \psi$. Combine these, we see the area of the "lune" is given by:

$$\Omega(\text{Lune}_{\hat{x}\hat{y}}(\rho)) = 2\pi - 2 ( 2\theta\rho + \pi - \psi ) = 2\psi - 4\theta\rho\\ = 2\cos^{-1}\left(\frac{\rho^2}{1-\rho^2}\right) - 4\rho\cos^{-1}\left(\frac{\rho}{\sqrt{1-\rho^2}}\right) $$

For $\rho$ between $\frac{1}{\sqrt{3}}$ and $\frac{1}{\sqrt{2}}$, the portion of the unit sphere outside the cube $[-\rho,\rho]^3$ is composed of 6 caps and these 6 caps intersect at 12 lunes. As a result, the surface area of the unit sphere inside the cube is given by

$$\begin{align} & 4\pi - 6\Omega(\text{Cap}(\rho)) + 12\Omega(\text{Lune}(\rho))\\ = & 12\pi(\rho-\frac{2}{3}) + 24\left[ \cos^{-1}\left(\frac{\rho^2}{1-\rho^2}\right) - 2\rho\cos^{-1}\left(\frac{\rho}{\sqrt{1-\rho^2}}\right)\right] \end{align}$$

Apply these to the problem of a sphere of radius $r$ and a co-centric unit cube, the value we should use for $\rho$ is $\frac{1}{2r}$. When $\frac{\sqrt{2}}{2} < r < \frac{\sqrt{3}}{2}$, the total area of the 8 "triangular shaped patches" of sphere inside the cube is given by:

$$6\pi r - 8\pi r^2 + 24 r\left[ r\cos^{-1}\left(\frac{1}{4r^2-1}\right) - \cos^{-1}\left(\frac{1}{\sqrt{4r^2-1}}\right) \right] $$ Projection of spherical Lune

Notes

$\color{blue}{[1]}$ Another way to derive the $2\theta\rho$ term is interpret it within the framework of Gauss Bonnet theorem. The "sum of exterior angles" corresponds to a line integral of geodesic curvature over corresponding edge. As a curve in $\mathbb{R}^3$, the edge of a lune has curvature $\frac{1}{\sqrt{1-\rho^2}}$ and its normal is making an angle $\sin^{-1}(\rho)$ with the normal of $S^2$. As a result, the geodesic curvature $\kappa_g$ on any edge is $\frac{\rho}{\sqrt{1-\rho^2}}$ and the line integral becomes:

$$\int \kappa_g ds = \left(\frac{\rho}{\sqrt{1-\rho^2}}\right) \left( 2\theta \sqrt{1-\rho^2} \right) = 2\theta\rho$$

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