© T Madas. r The volume and the surface area of a sphere are given by:

© T Madas

r

= 24S r

= 343V r

The volume and the surface area of a sphere are given by:

© T Madas

A tennis ball can be modelled as a sphere with diameter 8 cm

Calculate:

1. The volume of the ball, correct to 3 significant figures2. The surface area of the ball, correct to 3 significant figures

r

= 343V r

8 cm

4 cm

V =

43

x π x 43

= 24S r

V =

256π3

V =

268 [3 s.f.]cm3

S =

4 x π x 42

S =

64π

S =

201 [3 s.f.]cm2

© T Madas

Find the volume of this compound shape in terms of π

131 2 2r

Using Pythagoras Theorem:

r

212+ 213= Û2r 144+ 169= Û2r 144-169= Û

2r 25= Û=5r

5

© T Madas

131 2

5

213V r h= p

13

The volume of the cone:

´p 25´ 12´ ÛV =

13´p 25´ 12´ ÛV =

100´p ÛV =

=100pV

100V = p


© T Madas

131 2

5

343V r= p

43´

The volume of the semi-sphere:

p 35´12´ ÛV =

46´p 125´ ÛV =5006 ´p ÛV =

= 2503 pV

Volume of a sphere

2503V = p

100V = p


© T Madas


100p131 2

5

2503+ p

Total volume of the object

ÛV =

= 5503 pV250

3V = p

100V = p300

3p 250

3+ p ÛV =

5503p ÛV =

550 5503 3Note that =:p p

© T Madas

Calculate the surface area of a semi-sphere of radius 6 cm, correct to 3 significant figures.

6 cm

The surface area of a semi-sphere consists of:

the area of a circle

plus the curved surface area of a semi-sphere

S = 4πr 2

A = πr 2

The surface area of a sphere is given by:

The area of a circle is given by:

© T Madas

Calculate the surface area of a semi-sphere of radius 6 cm, correct to 3 significant figures.

6 cm

4x πx 62 x12

S = 4πr 2

A = πr 2


The area of a circle is given by:

=72π

π x 62 = 36π108π≈ 339 cm2

[3 s.f.]

© T Madas

One of the two sections of a sand timer can be thought of as a semi sphere connected to a cylinder which is in turn connected to a cone.All three solids share the same axis of symmetry.The radius of these solids is 3 cm and the heights of the cylinder and the cone are 9 cm and 3 cm respectively.

Calculate the volume of the sand timer in terms of π.

3

9

3

3

3

Measurements in cm

V = 43 πr 3

volume of a sphere

V = πr 2h

volume of a cylinder

V = 13 πr 2

volume of a cone

h

43

x π x 33 x 12

18π

πx 32x 9

81π

13

x π x 33x 3

27π

© T Madas

One of the two sections of a sand timer can be thought of as a semi sphere connected to a cylinder which is in turn connected to a cone.All three solids share the same axis of symmetry.The radius of these solids is 3 cm and the heights of the cylinder and the cone are 9 cm and 3 cm respectively.

Calculate the volume of the sand timer in terms of π.

3

9

3

3

3

Measurements in cm

43

x π x 33 x 12

18π

πx 32x 9

81π

13

x π x 33x 3

27π

126π

126π

252π ≈ 792 cm3

© T Madas

A cylindrical bucket of radius 9 cm contains some water.

A metal sphere is gently lowered into the water, raising its level by 6 cm which completely covers the sphere.

Calculate the radius of the sphere to the nearest cm.

© T Madas

A cylindrical bucket of radius 9 cm contains some water.

A metal sphere is gently lowered into the water, raising its level by 6 cm which completely covers the sphere.

Calculate the radius of the sphere to the nearest cm.

6 c

m

9 cm

r

V = 43 πr 3

volume of a sphere

V = πr 2h


πx 92x 6 43

x πx r 3=

486 = 4r 3

3

4r 3= 1458

r 3 = 364.5

r = 364.5 3

r ≈ 7 cm

© T Madas

The figure below shows the cross section of a cylinder of base radius a with 3 spheres fitting “snugly” inside it.

1. Calculate the volume of the cylinder in terms of a.2. Write down the volume of the three spheres.3. Hence calculate what fraction of the cylinder is empty

a

6a

V = 43 πr 3

volume of a sphere

V = πr 2h


Vc = πa 2x 6a

Vc = πa 36

V3s =43 πa 3x 3

V3s = πa 34

Vc =

V 3s =

Empty =

6πa 3

4πa 3

2πa 3

EmptyVc

= 2πa

3

6πa

3

=13

What fraction would it be if 5 spheres fitted snugly inside a cylinder with the appropriate dimensions?

© T Madas

A metal ball bearing can be modelled as a sphere of radius 2.5 cm.The ball bearing is to be melted and remoulded into the shape of a cone with a height of 4.8 cm. Calculate the radius of the base of the cone, in cm correct to 2 significant figures.

V = 43 πr 3

volume of the sphere

V = 13 πr 2

volume of the cone

h

= 43 x πx 2.53≈ 65.4498cm3

= 13 x πx r

2 ≈ 5.0265 cm3x 4.8 r 2

5.0265r 2= 65.4498

r 2= 13.0209

r = 13.0209

r = 3.6cm [2 s.f.]

© T Madas

A sphere with a radius of α cm has the same volume as a cone with a base radius of α cm.

Find the height of the cone, in terms of α.

V = 43 πα 3

volume of the sphere

V = 13 πα 2

volume of the cone

h

43 πα 3 = 1

3 πα 2h 43

α 3 = 13α 2h

43

α = 13 h 3x x3

4α =h

h = 4α , i.e. the height of the cone is 4 times the base radius

© T Madas

The Earth can be modelled as a sphere of radius 6 400

000 m and it is estimated that 360 000 000 000 000 m2 are covered by sea.

1. Write 360 000 000 000 000 in standard form.

2. Calculate the surface area of the Earth in m2, giving your answer in standard form correct to 3 significant figures.

3. Calculate what percentage of the Earth’s surface is covered by sea, giving your answer to the nearest percentage.

4x πx 64000002

S = 4πr 2


≈ m25.15x 1014S =

360 000 000 000 000 = 3.6

3.6 x 1014

5.15 x 1014x 100 ≈ 70%

x 1014

© T Madas. r The volume and the surface area of a sphere are given by:

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