Suppose that you wrap sphere in aluminum foil, much like how kisses, the candy, are wrapped. You would lay the sphere on a flat piece of aluminum foil, then you would collect the edges of the foil over the top of the sphere and cut off, or tie the excess(link to pictures provided below).

Now suppose that you to cut the sphere $h$ units from the bottom(diametrically opposite to the tied/cut end) so that the resulting shape would be spherical cap wrapped in aluminum foil. What would the area of the unwrapped and straightened foil from the spherical cap be in terms of the radius, $r$, of the sphere and the height, $h$, of the cap?

Here are some pictures of this being done a lemon if it helps.

Attempt: The lower bound for the area will be the surface area of the spherical cap(excluding the base) itself since the foil wrap the cap completely but still has creases and wrinkles on its surface, so $A > 2\pi h r$. I think an upper bound would be the area of a square with side length that is the perimeter of the dome(if I could call it that)(see this picture), because if you try to wrap a hemisphere with a foil of such dimensions you still have some excess foil left to cut. So, $A < 4\left(\displaystyle\int_{-r}^{-r+h}{\left. \sqrt{1+\frac{{{x}^{2}}}{{{r}^{2}}-{{x}^{2}}}}dx\right.}\right)^2$

  • $\begingroup$ That is approximately correct. Since there is no unique way to wrap it up, the approximation remains. $\endgroup$ – Narasimham Mar 5 '17 at 21:21
  • $\begingroup$ Instead of the cube you could consider a cylinder of base radius $\rho$ and height $r\arcsin(\rho/r)$ (the latter being half of what you call the perimeter of the dome). $\endgroup$ – Aretino Mar 6 '17 at 12:40
  • $\begingroup$ If your solid is the portion of the ball (of radius $r$ centered at the origin) where $z \geq h$, and if you use a disk of foil whose center lies at the center of the base of the sphere, the radius has to be at least $$R = \sqrt{r^{2} - h^{2}} + r\arccos\tfrac{h}{r},$$the sum of the radius of the base and the arc along the surface of the sphere, giving an area $\pi R^{2}$. Conversely, it appears a disk of this radius can be suitably creased in order to cover the truncated sphere. $\endgroup$ – Andrew D. Hwang Mar 6 '17 at 12:48
  • $\begingroup$ @Aretino Just noticed I said cube in the post, I meant a square with side length equal to the length of the arc along the surface of the cap. I don't see how a cylinder you described would work as a better upper bound. $\endgroup$ – E7_82_8E Mar 6 '17 at 22:48
  • $\begingroup$ @Narasimham Even if the there isn't a unique way to wrap it up, the area of any two wraps done the way described in the post should have be very close(right?). Isn't a better approximate possible? $\endgroup$ – E7_82_8E Mar 6 '17 at 22:58

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