Math .. help me please, I'm so lost?

QUESTION ELEVEN

a)

Volume of a sphere : (4πr^3) / 3

Volume of a hemisphere is of course the half : (2πr^3) / 3

Volume of a cone : b*h/3 ; with b the area of the base (a circle), and h the height

Area of circle : πr²

So the volume of the cone can be expressed as : πhr² / 3

The ratio of the hemisphere's volume to that of of the cone :

((2πr^3) / 3) / (πhr² / 3)

= (2πr^3) / (πhr²)

= (2r^3) / (hr²)

= 2r / h

b)

Volume of a cylinder : b*h

Like we did with the cone : πhr²

The ratio of the hemisphere's volume to that of of the cylinder :

((2πr^3) / 3) / (πhr²)

= ((2r^3) / 3) / (hr²)

= ((2r) / 3) / h

= (2r) / (3h)

QUESTION THIRTEEN

a)

Let us calculate the volume of the three balls :

Volume of a sphere : (4πr^3) / 3

So, volume of the three spheres : 4πr^3

Now we need that of the can :

Volume of a cylinder : πhr²

This particular one's height is 3 balls diameters (6r) : 6r * πr² = 6πr^3

The ratio balls' volume / can's volume :

(4πr^3) / (6πr^3)

= 4/6

= 2/3

The balls take 2/3 of the can's volume.

b)

A sphere's volume (its radius must be half the cube's side) : (4πr^3) / 3

A cube's volume : c^3

The cube fits the sphere, meaning c=2r, its volume is then : (2r)^3 = 8r^3

Ratio sphere's volume to that of the cube :

((4πr^3) / 3) / (8r^3)

= ((4π) / 3) / 8

= (π / 3) / 2

= π / 6

The sphere takes π / 6 of the cube's volume.

11)a. The hemisphere and the cone fits in exactly inside the cylinder ==> radius of cylinder = radius of the hemisphere = radius of the cone = height of the cone = height of the cylinder. (1)

let 'r' be the radius and 'h' be the height of the cylinder.

then vol. of the cone

= 1/3 * pi * r^2 * h

= 1/3 * pi * r^3 from (1) as r = h.

similarly,

vol. of the hemisphere

= 2/3 * pi * r^3

therefore, vol.hemisphere/vol.cone = 2 . (r^3 cancels out).

b.vol. of hemisphere = 2/3 * pi * r^3

vol. of cylinder = pi * r^2 * h = pi * r^3 from (1)

therefore the ratio is 2/3.

13. a. the three balls fits exactly in the cylinder ==> 6r = h (r=radius of the balls, h= height of the cylinder). (2)

vol. of the three balls = 3 * 4/3 * pi * r^3 = 4 * pi * r^3 .

vol of the cylinder = pi * r^2 * h = pi * r^2 * 6r = 6 * pi * r^3. from (2)

therefore the ratio is : 4/6 = 2/3 = .67 = 67% (other terms cancel)

b. sphere fits snugly in the cube ==> s = 2r (s=side of the cube, r= radius of the sphere)

vol. of the sphere = 4/3 * pi * r^3

vol of the cube = s^3 = (2r)^3 = 8r^3

therfore vol sphere/ vol cube = pi/6 ( after simplifications) = 52.35%

therefore vol. outside the sphere is 47.65%

hope this helps

:)

Regarding the 11th question,

Vol of hemisphere - (2/3)πr^3

Vol of cone - πr^2h/3

V of hemisphere : V of cone

(2/3)πr^3 : (1/3)πr^2h

2r: h

Regarding the 13th a. question:---

The radius of ball = r.

As the three balls exactly fit the cylindrical can, height of the cylinder = h = (2r + 2r + 2r ) = 6r

Volume of cylinder = πr^2h

= πr^2 (6r)

= 6πr^3

Volume of the 3 sphere balls = 3(4/3 πr^3) = 4 πr^3

The part of balls inside the cylinder is (4 πr^3) / (6 πr^3) = 2/3

Regarding 13th b. question: -----

Let the radius of sphere = r

So, the length of cube = a = 2r

Volume of cube = a^3 = (2r)^3 = 8r^3

Volume of sphere = 4/3 πr^3

Vcube - Vsphere = (8r^3 - 4/3πr^3) = (8-4.18)r^3 = 3.82r^3

Percentage unoccupied by sphere = (3.82r^3 / 8r^3)*100 = 0.4775*100 = 47.75%