1. Determine the mass of a hemispherical copper container whose external and internal radii are 12 cm and 10 cm.
Assuming that 1 cm3of copper weighs 8.9 g.
2. If the volume of a sphere is 566 cm3, find its radius.
3. A metal plumb bob comprises a hemisphere surmounted by a cone. If the diameter of the hemisphere and cone are each 4 cm and the total length is 5 cm, find its total volume.
4. A marquee is in the form of a cylinder surmounted by a cone. The total height is 6 m and the cylindrical portion has a height of 3.5 m, with a diameter of 15 m. Calculate
the surface area of material needed to make the marquee assuming 12% of the material is wasted in the process.
5. Determine (a) the volume and (b) the total surface area of the following solids:
(i) a cone of radius 8.0 cm and perpendicular height 10 cm
(ii) a sphere of diameter 7.0 cm (iii) a hemisphere of radius 3.0 cm
(iv) a 2.5 cm by 2.5 cm square pyramid of perpendic- ular height 5.0 cm
(v) a 4.0 cm by 6.0 cm rectangular pyramid of perpen- dicular height 12.0 cm
(vi) a 4.2 cm by 4.2 cm square pyramid whose sloping edges are each 15.0 cm
(vii) a pyramid having an octagonal base of side 5.0 cm and perpendicular height 20 cm.
6. The volume of a sphere is 325 cm3. Determine its diameter.
7. A metal sphere weighing 24 kg is melted down and recast into a solid cone of base radius 8.0 cm. If the density of the metal is 8000 kg/m3 determine (a) the diameter of the metal sphere and (b) the perpendicular height of the cone, assuming that 15% of the metal is lost in the process.
8. Find the volume of a regular hexagonal pyramid if the perpendicular height is 16.0 cm and the side of base is 3.0 cm.
9. A buoy consists of a hemisphere surmounted by a cone.
The diameter of the cone and hemisphere is 2.5 m and the slant height of the cone is 4.0 m. Determine the volume and surface area of the buoy.
10. A petrol container is in the form of a central cylindrical portion 5.0 m long with a hemispherical section sur- mounted on each end. If the diameters of the hemisphere and cylinder are both 1.2 m determine the capacity of the tank in litres (1 litre=1000 cm3).
11. Figure 24.9 shows a metal rod section. Determine its volume and total surface area.
1.00 cm radius
2.50 cm
1.00 m
Fig. 24.9
186 Basic Engineering Mathematics
12. Find the volume (in cm3) of the die-casting shown in Fig. 24.10. The dimensions are in millimetres.
50 100
25
30 rad 60
Fig. 24.10
13. The cross-section of part of a circular ventilation shaft is shown in Fig. 24.11, ends AB and CD being open. Calculate (a) the volume of the air, correct to the nearest litre, contained in the part of the system shown, neglecting the sheet metal thickness, (given 1 litre=1000 cm3), (b) the cross-sectional area of the sheet metal used to make the system, in square metres, and (c) the cost of the sheet metal if the material costs
£11.50 per square metre, assuming that 25% extra metal is required due to wastage.
2 m
1.5 m
1.5 m
800 mm 500 mm
A
B
D C
Fig. 24.11
24.4 Volumes and surface areas of frusta of pyramids and cones
The frustum of a pyramid or cone is the portion remaining when a part containing the vertex is cut off by a plane parallel to the base.
Thevolume of a frustum of a pyramid or coneis given by the volume of the whole pyramid or cone minus the volume of the small pyramid or cone cut off.
Thesurface area of the sides of a frustum of a pyramid or coneis given by the surface area of the whole pyramid or cone minus the surface area of the small pyramid or cone cut off. This gives the lateral surface area of the frustum. If the total surface area of the frustum is required then the surface area of the two parallel ends are added to the lateral surface area.
There is an alternative method for finding the volume and surface area of afrustum of a cone. With reference to Fig. 24.12:
Volume=13πh(R2+Rr+r2) Curved surface area=πl(R+r)
Total surface area=πl(R+r)+πr2+πR2
r I h
R Fig. 24.12
Problem 16. Determine the volume of a frustum of a cone if the diameter of the ends are 6.0 cm and 4.0 cm and its perpendicular height is 3.6 cm.
Method 1
A section through the vertex of a complete cone is shown in Fig. 24.13.
Using similar triangles AP DP=DR
BR
Hence AP
2.0=3.6 1.0 from which AP=(2.0)(3.6)
1.0 =7.2 cm
The height of the large cone=3.6+7.2=10.8 cm.
Volumes of common solids 187
4.0 cm
2.0 cm
1.0 cm 3.0 cm
6.0 cm
3.6 cm
Q P A
E
C D
B R
Fig. 24.13
Volume of frustum of cone
=volume of large cone−volume of small cone cut off
=13π(3.0)2(10.8)−13π(2.0)2(7.2)
=101.79−30.16=71.6 cm3 Method 2
From above, volume of the frustum of a cone
=13πh(R2+Rr+r2), where R=3.0 cm, r=2.0 cm and h=3.6 cm
Hence volume of frustum
=13π(3.6)
(3.0)2+(3.0)(2.0)+(2.0)2
=13π(3.6)(19.0)=71.6 cm3
Problem 17. Find the total surface area of the frustum of the cone in Problem 16.
Method 1
Curved surface area of frustum=curved surface area of large cone−curved surface area of small cone cut off.
From Fig. 24.13, using Pythagoras’ theorem:
AB2=AQ2+BQ2, from which AB =
[10.82+3.02]=11.21 cm and
AD2=AP2+DP2, from which AD =
[7.22+2.02]=7.47 cm Curved surface area of large cone
=πrl=π(BQ)(AB)=π(3.0)(11.21)=105.65 cm2 and curved surface area of small cone
=π(DP)(AD)=π(2.0)(7.47)=46.94 cm2 Hence, curved surface area of frustum=105.65−46.94
=58.71 cm2 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=99.6 cm2
Method 2
From page 186, total surface area of frustum
=πl(R+r)+πr2+πR2,
where l=BD=11.21−7.47=3.74 cm, R=3.0 cm and r= 2.0 cm.
Hence total surface area of frustum
=π(3.74)(3.0+2.0)+π(2.0)2+π(3.0)2=99.6 cm2
Problem 18. A storage hopper is in the shape of a frustum of a pyramid. Determine its volume if the ends of the frustum are squares of sides 8.0 m and 4.6 m, respectively, and the perpendicular height between its ends is 3.6 m.
The frustum is shown shaded in Fig. 24.14(a) as part of a com- plete pyramid. A section perpendicular to the base through the vertex is shown in Fig. 24.14(b).
By similar triangles:CG BG=BH
AH HeightCG=BG
BH AH
=(2.3)(3.6)
1.7 =4.87 m Height of complete pyramid=3.6+4.87=8.47 m Volume of large pyramid=1
3(8.0)2(8.47)=180.69 m3
188 Basic Engineering Mathematics
4.6 cm
4.6 cm
8.0 m
8.0m
2.3 m 2.3 m 3.6 m
4.0 m 2.3 m
1.7 m
(a) (b)
C
G D
B
A H
E F
Fig. 24.14
Volume of small pyramid cut off
= 1
3(4.6)2(4.87)=34.35 m3 Hence volume of storage hopper
=180.69−34.35=146.3 m3
Problem 19. Determine the lateral surface area of the storage hopper in Problem 18.
The lateral surface area of the storage hopper consists of four equal trapeziums.
4.6 m 4.6 m
8.0 m 0
8.0 m
Q
T P
R
S
U
Fig. 24.15
From Fig. 24.15, area of trapeziumPRSU
= 12(PR+SU)(QT)
OT=1.7 m (same asAHin Fig. 24.14(b) andOQ=3.6 m.
By Pythagoras’ theorem, QT=
(OQ2+OT2)=
[3.62+1.72]=3.98 m Area of trapeziumPRSU=12(4.6+8.0)(3.98)=25.07 m2 Lateral surface area of hopper=4(25.07)=100.3 m2
Problem 20. A lampshade is in the shape of a frustum of a cone. The vertical height of the shade is 25.0 cm and the diameters of the ends are 20.0 cm and 10.0 cm, respect- ively. Determine the area of the material needed to form the lampshade, correct to 3 significant figures.
The curved surface area of a frustum of a cone
=πl(R+r) from page 186.
Since the diameters of the ends of the frustum are 20.0 cm and 10.0 cm, then from Fig. 24.16,
r=5.0 cm, R=10.0 cm and l=
[25.02+5.02]=25.50 cm, from Pythagoras’ theorem.
5.0 cm r⫽5.0 cm
h⫽25.0 cm
R⫽10.0 cm I
Fig. 24.16
Hence curved surface area=π(25.50)(10.0+5.0)=1201.7 cm2, i.e. the area of material needed to form the lampshade is 1200 cm2, correct to 3 significant figures.
Problem 21. A cooling tower is in the form of a cylinder surmounted by a frustum of a cone as shown in Fig. 24.17.
Determine the volume of air space in the tower if 40% of the space is used for pipes and other structures.
12.0 m
25.0 m
12.0m 30.0m
Fig. 24.17
Volumes of common solids 189
Volume of cylindrical portion
=πr2h=π 25.0
2 2
(12.0)=5890 m3 Volume of frustum of cone
= 1
3πh(R2+Rr+r2)
where h=30.0−12.0=18.0 m, R=25.0/2=12.5 m and r=12.0/2=6.0 m.
Hence volume of frustum of cone
= 1 3π(18.0)
(12.5)2+(12.5)(6.0)+(6.0)2
=5038 m3 Total volume of cooling tower=5890+5038=10 928 m3. If 40% of space is occupied then volume of air space
=0.6×10 928=6557 m3
Now try the following exercise