Intriguing Ornament (the End)

Here's a better picture, showing the sphere is actually larger than shown.

(In fact, Otto, if you want to use this picture instead, you may. I could tone down the colors if you'd like.)

I found the surface area by projecting the sphere out to a cylinder and using Archimedes' Hat-Box Theorem

Let $x$ be the height along the axis of the cylinder. With $x=0$ the center, $x=-3$ the bottom, and $x=3$ the top. The regions where the sphere is pierced by the cube correspond to the intervals $[-2,-1]$ and $[1,2]$

Let's look at cross sections in these intervals

On the left is a side view of the ornament with the vertical axis the axis of the cylinder. The dashed line is a cross section at height $x$ and to the right is this top view of this section. The radius of this section is $\sqrt{9-x^{2}}$ and the angle drawn is $\arccos \frac{2}{\sqrt{9-x^{2}}}$ .

The portion of a circle is $8$ of these angles and the radius of the cylinder is 3, so the arc length at height $x$ is given by $24 \cdot \arccos \frac{2}{\sqrt{9-x^{2}}}$ .

$\int_{1}^{2} 24\cdot \arccos \frac{2}{\sqrt{9-x^{2}}} \approx 16.117$

The unobstructed portions of the sphere are 4 units along the axis so have area $4\cdot 6 \pi = 24 \pi$

The total area is then $2 \cdot 16.117 + 24 \pi = 107.632$ so the final answer is $\boxed{108} cm^{2}$

As much as I am a fan of brute-force methods, I look forward to seeing more elegant solutions to this one...

We have to first be aware that the given figure does not accurately depict the object described in the problem. Since $a = 4$ and $R = 3$ , most of the exposed area is the sphere's. Therefore, a reasonable approach would be to subtract the sphere's unexposed areas from $4\pi R^2 = 36\pi$ . By symmetry, the unexposed area is $8$ times the area of the sphere that is eclipsed by the cube (see figure). So $A_{\text{unexposed}} = 8\iint_{\text{eclipsed}} R^2 \sin\theta\,d\theta\,d\phi$ .

Most of the work lies in figuring out the bounds for $\theta$ and $\phi$ . From the "side view," it is not so difficult to get $\arccos\left(\frac{2}{3}\right) \leq \theta \leq \arccos\left(\frac{1}{3}\right)$ . From "top view," we find that how much $\phi$ is swept depends on $\theta$ as: $0 \leq \phi \leq \frac{\pi}{2} - 2\arccos\left(\frac{2}{3\sin\theta}\right)$ . (*)

So, $A_{\text{exposed}} = 36\pi - A_{\text{unexposed}} = 36 \pi - 8\int_{\arccos\left(\frac{2}{3}\right)}^{\arccos\left(\frac{1}{3}\right)}\int_{0}^{\frac{\pi}{2} - 2\arccos\left(\frac{2}{3\sin\theta}\right)} 3^2\sin\theta\,d\phi\,d\theta \approx 107.632 \approx \boxed{108}$ .

(*) At this step it's probably most helpful to make a sketch and see that $\phi$ is just $\frac{\pi}{2}$ minus the "other two angles," which are the same and equal to $\arccos\left( \frac{2}{3\sin\theta} \right)$ .

Comrade Huan has written a very careful solution, using spherical coordinates. For the sake of variety, let me propose a solution based on Cartesian coordinates; I find it easier to figure out the limits of integration that way (but this is a matter of taste and habit).

We can start by observing that each of the six caps (where the absolute value of one of the variables is $\geq 2$ ) has a surface area of $S_{cap}=2\pi Rh=6\pi$ . But we cannot simply add those surface areas up since, along each of the 12 edges, the two adjacent caps will intersect in a "lune," as Huan's figure illustrates. Focusing on the lune $y\geq 2$ and $z\geq 2,$ the intersection of the caps $z \geq 2$ and $y\geq 2$ , we find $S_{lune}=\int_{-1}^1\int_{2}^{\sqrt{5-x^2}}\frac{3}{\sqrt{9-x^2-y^2}}\ dy\ dx$ $\approx 0.455477$ . The surface area we seek is $6 S_{cap}-12 S_{lune}=60\pi-264\arctan(2)+72\arctan(\frac{11}{2})+240\arctan(\frac{1}{2}) \approx 107.63\approx \boxed{108}$ . (This is a primitive case of the principle of inclusion-exclusion.)