sphere and inscribed cube

Question

We can express ratio between the volume of a cube and volume of inscribed sphere in this cube as $6/\pi$ . What is the ratio between these objects when the cube is inscribed in a sphere?

Answer 1

Consider the cube within the sphere. It has sides $s$ and long diagonal $2a$, where $a$ is the radius of the sphere. Applying the Pythagorean theorem twice, we can show that $s=2a/\sqrt{3}$. It then follows that

$$\frac{V_{sphere}}{V_{cube}}=\frac{\frac{4}{3}\pi a^3}{\frac{8a^3}{3\sqrt{3}}}=\frac{\sqrt{3}\pi}{2}\approx 2.7207$$

Answer 2

As the cube is inscribed, all its vertices will be touching the sphere. Thus, the largest diagonal of the cube will be equal to the diameter of the sphere.

If we let the edge length of the cube be $a$, then the largest diagonal of the cube will be equal to $a\sqrt3$ by the Pythagorean Theorem, and this will also be the diameter.

The volume of the cube will be $a^3$

From the formula for the volume of a sphere given the diameter ($\frac{1}{6}\pi d^3$), the area of the sphere is equal to $\frac{\pi}{6}\times (\frac{a\sqrt3}{2})^3=\frac{\pi a^3\sqrt3}{16}$.

Therefore, our ratio is $a^3/\frac{\pi a^3\sqrt3}{16}=\frac{16}{\pi\sqrt3}$

Answer 3

I've calculated, sphere:incribed cube $= 2.7206990463:1.$ Since its about ratio, we can make the diameter of the sphere anything, so I took $2.$ This makes the space diagonal of the cube$=2.$ The sides of the cube are thus $2/\sqrt{3}.$ The volume of cube$= 8\sqrt{3}/9.$ The volume of the sphere is $4(\pi)r^3/3$ and in this case, $4(\pi)/3.$ If you take these two numbers $(4(\pi)/3$ and $8\sqrt{3}/9),$ the ratio of sphere to inscribed cube is $2.7206990463:1.$

Answer 4

when the cube is inscribed in the sphere then all its vertices must lie on the sphere, therefore they are equidistant from the centre of the sphere. note that the centre of the cube(intersection of cube diagonals) is also equidistant from the vertices therefore centre of the cube and sphere match each other. therefore diagonal of the cube is the diameter of the sphere, then an easy Pythagorean theorem follows which connects the radius of the sphere and the side of the cube to each other.