Cube Sphere Intersection, Part II

Comrade Huan has written an elegant solution in spherical coordinates. For the sake of variety, let me propose a solution in Cartesian coordinates.

As in an earlier problem, I will rearrange things a bit to make it easier to "wrap my head around" this problem. Let's place the center of the sphere at $(0,0,0)$ and the center of the cube at $(0.5,1.5,1)$ (I'm permuting the coordinates). The surface $S'=C\cap S$ whose area we seek is the graph of the function $z=\sqrt{4-x^2-y^2}$ on the region $D$ given by $1.5\geq x \geq -0.5$ and $\sqrt{4-x^2}\geq y \geq 0.5$ . With the scaling factor being $||\vec{N}||=\frac{2}{\sqrt{4-x^2-y^2}}$ , the area we seek is $A=\int_{-0.5}^{1.5}\int_{0.5}^{\sqrt{4-x^2}}\frac{2}{\sqrt{4-x^2-y^2}}\ dy\ dx\approx 5.1674$ The required answer is $\boxed{5167}$ .

I performed a single (but rather painstaking) computation on my trusted Wolfram's Mathematica program:

$A = \int_{\theta = \cos^{-1}\left( \frac{1}{4}\right) }^{\theta= \pi - \cos^{-1}\left(\frac{3}{4} \right) } \int_{\phi = \pi}^{\phi = \pi + \cos^{-1}\left( \frac{1/2}{2\sin\theta}\right) } (2^2)\sin(\theta)\,d\phi\,d\theta \approx 5.167$ .

So, the answer is $1000A \approx \boxed{5167}$ .

---

Most of the work lies in figuring out the bounds on $\theta$ and $\phi$ in the integral. $0 \leq \theta \leq \pi$ is the angle measured from the $z'$ -axis, and $0 \leq \phi \leq 2\pi$ is the angle measured from the $x'$ -axis in the $x'y'$ -plane, where the $(x',y',z')$ is a new coordinate system such that the sphere is centered at $(0,0,0)$ . It is not hard to see that the sphere is centered at $\left( \frac{3}{2}, 1, \frac{1}{2} \right)$ in $(x,y,z)$ , and that the top and bottom of the cube intersect the sphere at $z = -1$ and $z = +1$ , respectively. This means, in $(x',y',z')$ , $\cos^{-1}\left( \frac{1}{4}\right) \leq \theta \leq \pi - \cos^{-1}\left(\frac{3}{4} \right)$ .

It's a little trickier to find the bounds for $\phi$ , since we first have to realize the bounds of $\phi$ depend on the latitude of the intersection, or the angle $\theta$ . From top view, $\pi \leq \phi = \cos^{-1}\left( \frac{1/2}{2\sin\theta}\right)$ . Therefore, we get $\pi \leq\phi\leq \pi + \cos^{-1}\left( \frac{1/2}{2\sin\theta}\right)$ .

Lastly, we know that the area element of a sphere of radius $R$ is $dA = R^2\sin\theta\,d\phi\,d\theta$ , then put everything together to get the above integral and compute.