Basic Engineering Mathematics, Fifth Edition

Volumes of common solids 247

3. A sphere has a diameter of 6cm. Determine
   its volume and surface area.


4. If the volume of a sphere is 566cm^3 , find its
   radius.


5. A pyramid having a square base has a per-
   pendicular height of 25cm and a volume of
   75cm^3. Determine, in centimetres, thelength
   of each side of the base.


6. A cone has a base diameter of 16mm and a
   perpendicular height of 40mm. Find its vol-
   ume correct to the nearest cubic millimetre.


7. Determine (a) the volume and (b) the surface
   area of a sphere of radius 40mm.


8. The volume of a sphere is 325cm^3. Deter-
   mine its diameter.


9. Given the radius of the earth is 6380km,
   calculate, in engineering notation
   (a) its surface area in km^2.

(b) its volume in km^3.

10. An ingot whose volume is 1.5m^3 is to be
    made into ball bearings whose radii are
    8 .0cm. How many bearings willbe produced
    from the ingot, assuming 5% wastage?

27.3 Summary of volumes and surface

areas of common solids

A summary of volumes and surface areas of regular
solids is shown in Table 27.1.

Table 27.1 Volumes and surface areas of regular
solids

Rectangular prism

(or cuboid)

h

b

l Volume=l×b×h

Surface area= 2 (bh+hl+lb)

Cylinder

h

r

Volume=πr^2 h

Total surface area= 2 πrh+ 2 πr^2

Triangular prism

I

b

h

Volume=

1

2

bhl

Surface area=area of each end+

area of three sides

Pyramid

h

A

Volume=

1

3

×A×h

Total surface area=

sum of areas of triangles

forming sides+area of base

Cone

h

r

l

Volume=

1

3

πr^2 h

Curved surface area=πrl

Total surface area=πrl+πr^2

Sphere

r Volume=

4

3

πr^3

Surface area= 4 πr^2

27.4 More complex volumes and surface areas

Here are some worked problems involving more com-

plex and composite solids.

Problem 16. A wooden section is shown in

Figure 27.14. Find (a) its volume in m^3 and

(b) its total surface area