# Cube and Spheres

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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*

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Fixed version:

Spoiler

Volume of cube is

$x^3$

The volume of a sphere inscribed in a cube is:

$\frac{4}{3} \pi \frac{x}{2}^3$

The volume of a sphere superscribed is:

$\frac{4}{3} \pi \sqrt{ 3 \frac{x}{2}^2}^3$

4 hours ago, Commander said:

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

$\frac{\frac{4}{3} \pi (\frac{x}{2})^3}{x^3}=\frac{\pi}{6}= 52.4\%$

4 hours ago, Commander said:

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

$\frac{ \frac{4}{3} \pi \sqrt{ 3 \frac{x}{2}^2}^3 }{\frac{4}{3} \pi \frac{x}{2}^3}=5.2$

Edited by Sensei
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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.

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This may be a shortcut?

Spoiler

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

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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?

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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.

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4 minutes ago, Sensei said:
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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.

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Spoiler

So cylinder v at given height = hπx2

then v ball is 4/3π(x/2)3

h=x/2 + x/2 + x/2(21/2

(simple Pythagorean geometry)

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

Then h = 5+5+5(21/2)=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)

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8 minutes ago, TheVat said:
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So cylinder v at given height = hπx2

then v ball is 4/3π(x/2)3

h=x/2 + x/2 + x/2(21/2

(simple Pythagorean geometry)

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

Then h = 5+5+5(21/2)=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

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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 !

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