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© T Madas

Axis

Vertex

Generator

Radius

Base

SlantHeight

C y l i n d e r s a n d C o n e s

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Volume of a Cylinder

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The Volume of a Cylinder

2V r h=

r

h

A cylinder is a P

whose cross section is a c

rism

ircle

V =Base Areax Height

= πr 2 x h

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V =πr 2 h

V =πx 102 x 40

V ≈12566 cm3[n. w. n.]

x

V =x-sectional Area heightx

Calculate the volume of these cylinders

10 cm

40 c

m

V =πr 2 h

V =πx 52x 22

V ≈1728 m3[n. w. n.]

x

V =x-sectional Area heightx

5 m

22 m

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V =πr 2 h

V =πx 122 x 35

V ≈15834 cm3[n. w. n.]

x

Calculate the volume of these cylinders

12 cm

35 c

m

V =πr 2 h

V =πx 42x 18

V ≈905 m3[n. w. n.]

x

4 m

18 m

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Surface Area of a Cylinder

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2C r=

2A rh= h

2A r=

2A r=

The Surface Area of a Cylinder

22 2A rh r = +

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S = π r h

S =

S ≈3142 cm2[n. w. n.]

+

Calculate the Surface Area of these Cylinders

10 cm

40 c

m

5 m

22 m

2 π r 22

x πx 10x 40+2 x πx 1022

S ≈2513.27 628.32+

S = π r h

S =

S ≈ 848 m2[n. w. n.]

+2 π r 22

x π x 5x 22+2 x πx 522

S ≈691.15 157.08+

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S = π r h

S =

S ≈6126 cm2[n. w. n.]

+

Calculate the Surface Area of these Cylinders

15 cm

50 c

m

4 m

14 m

2 π r 22

x πx 15x 50+2 x πx 1522

S ≈4712.39 1413.72+

S = π r h

S =

S ≈ 452 m2 [n. w. n.]

+2 π r 22

x π x 4x 14+2 x πx 422

S ≈351.86 100.53+

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Volume of a Cone

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The Volume of a Cone

213

V r h= ´ ´h

It can be shown that for a cone:

V = Base Areax Height

r

13

´

213

V r h=

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13

x π x 42x 9

Calculate the volume of these cones

9 c

m

4 cm

8 c

m

6 cm

V = πr 2h13

=

151≈ cm3

13

x π x 62x 8

V = πr 2h13

=

302≈ cm3

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13

x π x 32x 12

Calculate the volume of these cones

12 c

m

3 cm

9 c

m

5 cm

V = πr 2h13

=

113≈ cm3

13

x π x 52x 9

V = πr 2h13

=

236≈ cm3

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h

8.8

4 c

m

A cylinder is shown below.The radius of its base is 6 cm and has a volume of 1000 cm3.Calculate its surface area to 3 significant figures.

6 cmVolume = base area x height

1000 = 6 x 6 x h

1000 ≈ 113.1 x h

h ≈ 1000 ÷ 113.1

h ≈ 8.84 cm

x π

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8.8

4 c

m

A cylinder is shown below.The radius of its base is 6 cm and has a volume of 1000 cm3.Calculate its surface area to 3 significant figures.

Surface Area:

6 x 6 ≈ 226.2 cm2x π6 cm

x 2

8.8

4 c

m

Circumference of circle

2 x 6 ≈ 333.3 cm2x π x 8.84

559.5 cm2

560 cm2 [ 3 s.f.]

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© T Madas

A tank without a lid is in the shape of a cylinder.The radius of its base is 30 cm and has a capacity of 225 litres.Calculate its surface area, to 2 significant figures.

h

79.6

cm

30 cm

Volume = base area x height

225000 = 30 x 30 x h

225000 ≈ 2827 x h

h ≈ 225000 ÷ 2827

h ≈ 79.6 cm

x π

1 litre = 1000 cm3

225 litres = 225000 cm3

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A tank without a lid is in the shape of a cylinder.The radius of its base is 30 cm and has a capacity of 225 litres.Calculate its surface area, to 2 significant figures.

h

79.6

cm

30 cm

79.6

cm

Circumference of circle

Surface Area:

30 x 30 ≈ 2827 cm2x π

2 x 30 ≈ 15004 cm2x π x 79.6

17831 cm2

18000 cm2 [ 2 s.f.]

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© T Madas

A can without a lid is in the shape of a cylinder.The radius of its base is 4 cm and has a capacity of 192π cm3.Calculate its surface area in terms of π.

h

12 c

m4 cm Volume = base area x height

192π = 4 x 4 x h

192π 16πh

h = 12 cm

x π

= 16π16π

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12 c

m

A can without a lid is in the shape of a cylinder.The radius of its base is 4 cm and has a capacity of 192π cm3.Calculate its surface area in terms of π.

4 cmSurface Area:

4 x 4 ≈ 16π cm2x π

2 x 4 ≈ 96π cm2x π x 12

112π cm2

12 c

m

Circumference of circle

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© T Madas

131 2 2r

Using Pythagoras Theorem:

r

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

2r 25= Û=5r

5

Find the volume of this compound shape in terms of π

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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

Find the volume of this compound shape in terms of π

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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

Find the volume of this compound shape in terms of π

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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

Find the volume of this compound shape in terms of π

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STOP

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L

r

Lh

r

r 2 r

θ

r 2 +h2 = L2

2 r

´ L ´

360´ 2 = r2

A

L2´

360=A

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r 2 +h2 = L2 ´ L ´

360´ 2 = r2

L2´

360=A

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´

360= r2´ 2 ´ L

r 2 +h2 = L2

L2´

360=A

Û

L2

360= r2 Û

L

360= r Û

=L r360 Û

=L

r360

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´r360

r 2 +h2 = L2

L2´

360=A

=L

r360

Û L2´360

=A L

r360

Û

L2 ´360

1

=A L

r360

Û

L2

L360=A Û

=LrA

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r 2 +h2 = L2

L2´

360=A

=L

r360

Û =L +r h2 2

=LrA

=A r +r h2 2

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=LrA

=A r +r h2 2

Lh

r

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