If *C* is a Cube and *S1* is the Sphere Inscribed and *S2* is the Sphere Superscribed then 
(1) What is the fraction of the volume of *C* is contained in *S1*
(2) How many times the volume of *S2* is that of *S1*

Fixed version:

Spoiler

Volume of cube is

[math]x^3[/math]

The volume of a sphere inscribed in a cube is:

[math]\frac{4}{3} \pi \frac{x}{2}^3[/math]

The volume of a sphere superscribed is:

[math]\frac{4}{3} \pi \sqrt{ 3 \frac{x}{2}^2}^3[/math]

4 hours ago, Commander said:

(1) What is the fraction of the volume of *C* is contained in *S1*

[math]\frac{\frac{4}{3} \pi (\frac{x}{2})^3}{x^3}=\frac{\pi}{6}= 52.4\%[/math]

  

4 hours ago, Commander said:

(2) How many times the volume of *S2* is that of *S1*

 

[math]\frac{ \frac{4}{3} \pi \sqrt{ 3 \frac{x}{2}^2}^3 }{\frac{4}{3} \pi \frac{x}{2}^3}=5.2[/math]

 

Edited September 10, 20232 yr by Sensei

Spoiler

Straightforward application of known formulas for volume of a cube and a sphere. Take a length of the cube edge to be 2a to simplify the algebra.

 

This may be a shortcut?

 

Spoiler

Just take diagonal of a unit cube, which is 1.73, and cube that, so 5.178

 

 

Let's make this a bit more interesting.

Consider a cylindrical glass and four identical balls packed into it compactly, as shown:

How much water is needed to exactly cover the balls compared to the volume occupied by the balls?

8 minutes ago, Genady said:

How much water is needed to exactly cover the balls compared to the volume occupied by the balls?

Spoiler

The answer depends on the density of the balls.. 😛

If their density is less than 1g/mL of water, they will float on the surface and will only be partially submerged.

 

4 minutes ago, Sensei said:

  Reveal hidden contents

The answer depends on the density of the balls.. 😛

If their density is less than 1g/mL of water, they will float on the surface and will only be partially submerged.

 

Spoiler

Correct. The answer also depends on their chemical composition.

 

Spoiler

let x=cylinder radius

and x/2 is ball radius

So cylinder v at given height = hπx²

then v ball is 4/3π(x/2)³

h=x/2 + x/2 + x/2(2^1/2) 

(simple Pythagorean geometry)

So then plug in a number for x, like 10.

Then h = 5+5+5(2^1/2)=17.07, ergo undisplaced volume is roughly

5363.

balls volume is

4 x (4/3π5³) = 4x523.6=2094

Then, take 5363  - 2094 = 3269

So....cylinder has 3269 units of submerging liquid with balls volume of 2094. (assuming balls denser than liquid)

Roughly 1.56.  (some rounding errors may have accrued)

 

 

 

 

 

8 minutes ago, TheVat said:

  Hide contents

let x=cylinder radius

and x/2 is ball radius

So cylinder v at given height = hπx²

then v ball is 4/3π(x/2)³

h=x/2 + x/2 + x/2(2^1/2) 

(simple Pythagorean geometry)

So then plug in a number for x, like 10.

Then h = 5+5+5(2^1/2)=17.07, ergo undisplaced volume is roughly

5363.

balls volume is

4 x (4/3π5³) = 4x523.6=2094

Then, take 5363  - 2094 = 3269

So....cylinder has 3269 units of submerging liquid with balls volume of 2094. (assuming balls denser than liquid)

Roughly 1.56.  (some rounding errors may have accrued)

 

 

 

 

 

Yes! +1

Yes TY all for answers !

Pi/6 is the Fraction and 3xSqr Rt 3 times is the Volume of the Bigger Sphere to that of the Smaller one !

Quite Straight forward !