Sphere in cubical box with same volume

Question

A sphere with a radius of one is put inside a cubical box with circular holes cut out such that the ball fits perfectly. The box and the sphere share the same volume; find the surface area of the box.

I tried cutting the cube down the middle to get a square and a circle. I then tried to find the distance between the two points of intersection between the circle and square on one side of the square but hit a dead end. I then tried to find the volume of empty space in the box since that could help me find the volume of the protruding parts of the sphere and maybe lead to the answer but it led to no avail.

Answer 1

The volume of the cube is

$ V_c = a^3 $

where $a$ is the side length of the cube.

The volume of the sphere is

$ V_s = \dfrac{4 \pi}{3} r^3 $

Setting $r = 1$ and equating volumes, we deduce that

$ a = \sqrt[3]{\dfrac{4\pi}{3}} \approx 1.61199195402$

Therefore, the radius of the circle cut from each of the six sides of the cube to fit the protruding sphere is

$ R = \sqrt{1 - \left( \dfrac{a}{2} \right)^2 } \approx 0.59192101251$

Hence the total area of the cubical box after removing these circular cuts is

$ \text{Area} = 6 ( a^2 - \pi R^2 ) \approx 8.98678 $

Answer 2

I got 6*(4pi/3)^(2/3) * (1+pi/4) - 6pi as the answer. I'm sorry I don't have much experience with LaTeX and all. But here's how I worked this out: I found the side of the square from the volume of the circle, and after that just deducted the circle's area from a square's, and the surface area follows as six times this.

Answer 3

Since the cube and sphere have the same volume it is easy enough to equate the two and solve for the side S of the cube.

Now referring to this sketch:

It is easy to see that one side of the red right angle triangle that is perpendicular to the face of the cube is equal to S/2 and the hypotenuse is the same as the radius (R=1) of the sphere.

Using Pythagoras theorem, $R^2 = R_1^2 + (S/2)^2$,

$\Rightarrow R_1 = \sqrt{R^2 - (S/2)^2} = \sqrt{1 - S^2/4}$

This is the radius of one of the cut out circles each with a area of $\pi R_1^2$. Subtract 6 of these circle areas from the surface area of the cube (6*S^2).

You could of course just measure $R_1$ on my sketch with a ruler. It's very accurately drawn :-P