Cube & Sphere, Sphere & Cube

Formula for the volume of a sphere: $V = \frac{4}{3} \pi r^3$ where $r$ is the radius of the sphere, or $V = \frac{\pi}{6} d^3$ where $d$ is the diameter of the sphere.

Sphere Inside A Cube

For a sphere inscribed in a cube, the diameter of the sphere is equal to the height of the cube. Therefore if the cube has volume, $V_{cube-outside} = x^3,$ then the height of the cube is $x$ . Therefore, the volume of the inscribed sphere is $V_{sphere-inside} = \frac{\pi}{6} x^3$ .

Cube Inside A Sphere

The trickier case is the cube inscribed inside the sphere. The key observation is that each line from the center of the cube to one of its 8 corners is a radius of the surrounding sphere. Or, in other words, a line between two opposite corners of the cube is the diameter of the surrounding sphere:

If the surrounding sphere has volume $V_{sphere-outside} = x^3,$ then the diameter of that sphere, $d$ is found as:

$V_{sphere-outside} = x^3 = \frac{\pi}{6} d^3$
$\frac{6}{\pi} x^3 = d^3$
$d = \sqrt[3]{\frac{6}{\pi} x^3}$

Now, let $d$ continue to be the diagonal between two opposite corners of the cube,
let $q$ be the diagonal of one square face of the cube, and
let $s$ be the length of each edge of the cube.

Then, using the Pythagorean theorem, we have that $q^2 + s^2 = d^2.$ Furthermore, since $s$ is the length of each edge of the cube, then $2s^2 = q^2.$ Therefore, $3s^2 = d^2,$ and, taking the square root of both sides, we get that $\sqrt{3}s = d$ , or, alternately, $s = \frac{d}{\sqrt{3}}.$

Finally, recall from above that, for a sphere of volume $x^3$ , we we found that $\frac{6}{\pi} x^3 = d^3$ , where $d$ is the diameter of the sphere and the diagonal across the inscribed cube. Therefore, we can now find that the volume of the inscribed cube. The side length of the cube is $s$ , therefore, the volume of the cube is $V_{cube-inside} = s^3 = \frac{d^3}{(\sqrt{3})^3} = \frac{6}{\pi3^{\left(\frac{3}{2}\right)}} x^3 = \frac{2}{\pi\sqrt{3}} x^3.$

Therefore, $V_{sphere-inside} = \frac{\pi}{6} x^3 > \frac{2}{\pi\sqrt{3}} x^3 = V_{cube-inside}.$
The volume of the grey sphere inscribed within the cube is greater than the volume of the grey cube inscribed within the sphere.

We assign to each of the white objects a volume of 1, and calculate the volume of the inscribed grey object. I will use the fact that the volume of a sphere is $V = \tfrac43\pi r^3 = \tfrac\pi 6 d^3$ .

Volume of the grey sphere:

We assume $V_{wc} = 1$ . The side of the white cube is therefore $a_{wc} = 1$ . This is equal to the diameter of the grey sphere: $d_{gs} = 1$ . Thus we find for the volume of the grey sphere, $V_{gs} = \frac\pi 6 d_{gs}^3 = \frac\pi 6 \approx 0.524.$

Volume of the grey cube:

We assume $V_{ws} = 1$ . The diameter of the white sphere is therefore $d_{ws} = \sqrt[3]{\frac 6\pi}.$ This is equal to the body diagonal of the grey cube. Therefore $\sqrt 3 a_{gc} = d_{ws} = \sqrt[3]{\frac 6 \pi}; \\ a_{gc} = \frac{\sqrt[3]{6/\pi}}{\sqrt 3}. \\ V_{gc} = a_{gc}^3 = \frac{6/\pi}{(\sqrt 3)^3} = \frac{2}{\sqrt 3\pi} \approx 0.368.$

Thus the grey sphere has a significantly greater volume than the grey cube.