Basic Engineering Mathematics, Fifth Edition

Volumes of common solids 253

4.0cm

P

E

A

D

B R

Q

C

2.0cm

3.6cm

1.0cm

3.0cm

6.0cm

Figure 27.22

The height of the large cone= 3. 6 + 7. 2

= 10 .8cm

Volume of frustum of cone

=volume of large cone

−volume of small cone cut off

=

1

3

π( 3. 0 )^2 ( 10. 8 )−

1

3

π( 2. 0 )^2 ( 7. 2 )

= 101. 79 − 30. 16 =71.6cm^3

(ii) Method 2

From above, volume of the frustum of a cone

=

1

3

πh(R^2 +Rr+r^2 )

where R= 3 .0cm,

r= 2 .0cm and h= 3 .6cm

Hence,volume of frustum

=

1

3

π( 3. 6 )

[

( 3. 0 )^2 +( 3. 0 )( 2. 0 )+( 2. 0 )^2

]

=

1

3

π( 3. 6 )( 19. 0 )=71.6cm^3

Problem 25. Find the total surface area of the

frustum of the cone in Problem 24.

(i) Method 1

Curved surface area of frustum=curved surface

area of large cone−curved surface area of small

cone cut off.

From Figure 27.22, using Pythagoras’ theorem,

AB^2 =AQ^2 +BQ^2

from which AB=

√[

10. 82 + 3. 02

]

= 11 .21cm

and AD^2 =AP^2 +DP^2

from which AD=

√[

7. 22 + 2. 02

]

= 7 .47cm

Curved surface area of large cone=πrl

=π(BQ)(AB)=π( 3. 0 )( 11. 21 )

= 105 .65cm^2

and curved surface area of small cone

=π(DP)(AD)=π( 2. 0 )( 7. 47 )= 46 .94cm^2

Hence, curved surface area of frustum

= 105. 65 − 46. 94

= 58 .71cm^2

Total surface area of frustum

=curved surface area

+area of two circular ends

= 58. 71 +π( 2. 0 )^2 +π( 3. 0 )^2

= 58. 71 + 12. 57 + 28. 27 =99.6cm^2

(ii) Method 2

From page 252, total surface area of frustum

=πl(R+r)+πr^2 +πR^2

wherel=BD= 11. 21 − 7. 47 = 3 .74cm,

R= 3 .0cmandr= 2 .0cm. Hence,

total surface area of frustum

=π( 3. 74 )( 3. 0 + 2. 0 )+π( 2. 0 )^2 +π( 3. 0 )^2

=99.6cm^2