Unizor - Geometry3D - Spheres - Problems 3

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Notes to a video lecture on http://www.unizor.com

Spheres - Problems 3

Just as a reminder, here are a few formulas related to spheres. They are all derived in the previous lectures and are helpful in solving these problems.
Volume of a sphere: 4πR³/3
Surface area of a sphere: 4πR²
Volume of a cap: πH²(3R−H)/3
Surface area of a dome: 2πRH
Volume of a sector: 2πR²H/3

Problem A
Diameter of Jupiter is 11 times greater than diameter of Earth.
Diameter of Mars is half of diameter of Earth.
What is the ratio of surface and volume of Jupiter to those of Mars, assuming all the planets are spherical?

Answer:
The ratio of the surface areas of Jupiter to that of Mars is 484:
SJupiter = 484·SMars
The ratio of their volumes is 10648:
VJupiter = 10648·VMars

Problem B
A cylindrical glass of radius R is filled with water to the middle.
A ball (heavier than water) of a radius r (smaller than R) is dropped into this glass.
By how much the height of the water in a glass will rise?

Answer:
The water will rise by
h = 4r³/(3R²)

Problem C
An air-filled ball made of some heavier than water material is floating in the water such that exactly half of it is above the water.
According to Archimedes' principle of flotation, the weight of a ball equals to the weight of the water it displaced.
The ball's material unit weight is B, the water's unit weight is W.
Assume that the air inside the ball is weightless.
The radius of a ball is R
Construct the equation to solve to determine the width of the ball's wall.

Solution:
Assume that the width of the ball's wall is X. Then the internal radius of the ball is R−X.
The ball's weight equals to its unit weight B multiplied by the volume occupied by its wall, which is a difference between the volumes of two spheres, one has an external diameter of a ball, another is inside it and has its internal diameter:
B·[4πR³/3−4π(R−X)³/3]=
= 4πB[R³−(R−X)³]/3
This weight should be equal to the weight of displaced water, which has a unit weight W and the volume equal to half of the volume of a ball:
W·2πR³/3
The above two weights, that of the ball and that of the displaced water, must be equal, which produces the following equation for X:
4πB·[R³−(R−X)³]/3 =
= 2πW·R³/3
Simplifying this equation, we get
2B·[R³−(R−X)³] = W·R³
Generally speaking, this is an equation of the third degree and can be solved numerically.

Problem D
A sphere is inscribed into a regular tetrahedron, it's tangential to all its four faces.
Prove that the radius of this sphere is 1/4th of the altitude of a tetrahedron.