PROBLEM 5.38
Determine the volume and the surface area of the solid obtained
by rotating the area of Problem 5.4 about (a) the line x = 240 mm,
(b) the y axis.
SOLUTION
From the solution to Problem 5.4 we have
32
63
63
15.3 10 mm
2.6865 10 mm
1.4445 10 mm
A
xA
yA
= ×
Σ= ×
Σ= ×
Applying the theorems of Pappus-Guldinus we have
(a) Rotation about the line
240 mmx=
36
Volume 2 (240 )
2 (240 )
2 [240(15.3 10 ) 2.6865 10 ]
xA
A xA
π
π
π
= −
= −Σ
= ×− ×
63
Volume 6.19 10 mm
= ×
line line
1133445566
Area 2 2 ( )
2( )
XL x L
xL xL xL xL xL
ππ
π
= = Σ
= ++++
Where
16
,,xx
are measured with respect to line
240 mm.x=
Area 2 [(120)(240) (15)(30) (30)(270)
(135)(210) (240)(30)]
π
= ++
++
32
Area 458 10 mm= ×
(b) Rotation about the y axis
area
63
Volume 2 2 ( )
2 (2.6865 10 mm )
X A xA
ππ
π
= = Σ
= ×
63
Volume 16.88 10 mm= ×
line line
1122334455
Area 2 2 ( )
2( )
2 [(120)(240) (240)(300)
(225)(30) (210)(270) (105)(210)]
XL x L
xL xL xL xL xL
ππ
π
π
= = Σ
= ++++
= +
++ +
62
Area 1.171 10 mm= ×
consent of McGraw–Hill Education.
PROBLEM 5.39
Determine the volume and the surface area of the solid obtained by rotating the
area of Problem 5.8 about (a) the x axis, (b) the y axis.
SOLUTION
From the solution to Problem 5.8 we have
2
3
3
1146.57 in.
14147.0 in.
26897 in.
A
xA
yA
=
Σ=
Σ=
Applying the theorems of Pappus-Guldinus we have
(a) Rotation about the x axis:
3
Volume 2 2
2 (26897 in. )
area
Y A yA
ππ
π
= = Σ
=
or
33
Volume 169.0 10 in= ×
2233445566
Area 2
2( )
2( )
2 [(7.5)(15) (30)( 15) (47.5)(5)
(50)(30) (25)(50)]
line
line
YA
yA
yL yL yL yL yL
π
π
π
ππ
=
= Σ
= ++++
= + ×+
++
or
32
Area 28.4 10 in= ×
(b) Rotation about the y axis
3
Volume 2 2
2 (14147.0 in. )
area
X A xA
ππ
π
= = Σ
=
or
33
Volume 88.9 10 in= ×
1122334455
Area 2 2 ( )
2( )
2 15
2 (15)(30) (30)(15) 30 ( 15) (30)(5) (15)(30)
line line
XL x L
xL xL xL xL xL
ππ
π
ππ
π
= = Σ
= ++++
×
= + + − ×+ +
or
32
Area 15.48 10 in= ×
consent of McGraw–Hill Education.
PROBLEM 5.40
Determine the volume of the solid generated by rotating the
parabolic area shown about (a) the x-axis, (b) the axis AA′.
SOLUTION
First, from Figure 5.8a, we have
4
3
2
5
A ah
yh
=
=
(a) Rotation about the x–axis:
Volume 2
24
253
yA
h ah
π
π
=
=
or
2
16
Volume 15 ah
π
=
(b) Rotation about the line
:AA′
Volume 2 (2 )
4
2 (2 ) 3
aA
a ah
π
π
=
=
or
2
16
Volume 3ah
π
=
consent of McGraw–Hill Education.
PROBLEM 5.41
Determine the capacity, in liters, of the punch bowl
shown if R = 250 mm.
SOLUTION
The volume can be generated by rotating the triangle and circular sector shown about the y-axis. Applying
the second theorem of Pappus-Guldinus and using Figure 5.8a, we have
11 2 2
2
6
33
3
3
3
22
2( )
1 1 1 1 3 2 sin30
232 22 2 3 6
2163 23
33
8
33 (0.25 m)
8
0.031883 m
V xA xA
xA xA
R
R RR R
RR
R
π
ππ
π
π
π
π
π
π
= = Σ
= +
°
= × ×× +
×
= +
=
=
=
Since
33
10 l 1 m=
3
33
10 l
0.031883 m 1m
V= ×
31.9 litersV=
consent of McGraw–Hill Education.
PROBLEM 5.42
The aluminum shade for the small high–intensity lamp shown has a uniform thickness of 1 mm. Knowing
that the density of aluminum is 2800 kg/m3, determine the mass of the shade.
SOLUTION
The mass of the lamp shade is given by
m V At
ρρ
= =
where A is the surface area and t is the thickness of the shade. The area can be generated by rotating the line
shown about the x-axis. Applying the first theorem of Pappus Guldinus, we have
11 22 33 44
22
2( )
A yL yL
yL yL yL yL
ππ
π
= = Σ
= +++
or
22
22
22
2
13 mm 13 16
2 (13 mm) mm (32 mm) (3 mm)
22
16 28 mm (8 mm) (12 mm)
2
28 33 mm (28 mm) (5 mm)
2
2 (84.5 466.03 317.29 867.51)
10,903.4 mm
A
π
π
+
= + ×+
+
+ ×+
+
+ ×+
= +++
=
Then
3 32
(2800 kg/m )(10.9034 10 m )(0.001 m)
m At
ρ
−
=
= ×
or
0.0305 kgm=
consent of McGraw–Hill Education.
PROBLEM 5.43
Knowing that two equal caps have been removed from a 10–in.–diameter
wooden sphere, determine the total surface area of the remaining portion.
SOLUTION
The surface area can be generated by rotating the line shown about the y-axis. Applying the first theorem of
Pappus-Guldinus, we have
11 2 2
22
2 (2 )
A XL x L
xL xL
ππ
π
= = Σ
= +
Now
4
tan 3
a
=
or
53.130
a
= °
Then
2
180
5 in. sin53.130
53.130
4.3136 in.
x
π
°
×°
=°×
=
and
2
2 53.130 (5 in.)
180
9.2729 in.
3
2 2 in. (3 in.) (4.3136 in.)(9.2729 in.)
2
L
A
π
π
= °×
°
=
= +
or
2
308 inA=
consent of McGraw–Hill Education.
PROBLEM 5.44
Three different drive belt profiles are to be
studied. If at any given time each belt
makes contact with one–half of the
circumference of its pulley, determine the
contact area between the belt and the
pulley for each design.
SOLUTION
Applying the first theorem of Pappus-Guldinus, the contact area
C
A
of a
belt is given by
C
A yL yL
ππ
= = Σ
where the individual lengths are the lengths of the belt cross section that
are in contact with the pulley.
(a)
11 2 2
[2( ) ]
0.125 0.125 in.
2 3 in. [(3 0.125) in.](0.625 in.)
2 cos20
C
A yL yL
π
π
= +
= − +−
°
or
2
8.10 in
C
A=
(b)
11
[2( )]
0.375 0.375 in.
2 3 0.08 in.
2 cos20
C
A yL
π
π
=
= −−
°
or
2
6.85 in
C
A=
(c)
11
[2( )]
2(0.25)
3 in. [ (0.25 in.)]
C
A yL
π
ππ
π
=
= −
or
2
7.01in
C
A=
consent of McGraw–Hill Education.
PROBLEM 5.45
Determine the volume and weight of the solid brass knob shown, knowing
that the specific weight of brass is 0.306 lb/in3.
SOLUTION
Area, in2
, in.y
3
, inyA
1
2
(0.75) 0.4418
4
π
=
0.8183 0.3615
2
(0.5)(0.75) 0.375=
0.25 0.0938
3
(1.25)(0.75) 0.9375=
0.625 0.5859
4
2
(0.75) 0.4418
4
π
−= −
0.9317 −0.4116
Σ
0.6296
33
2 2 (0.6296 in ) 3.9559 inV yA
ππ
=Σ= =
3
3.96 inV=
33
(0.306 lb/in )(3.9559 in )WV
γ
= =
1.211lbW=
Volume of knob is obtained by rotating area
of components shown below.
consent of McGraw–Hill Education.
PROBLEM 5.46
Determine the total surface area of the solid brass knob shown.
SOLUTION
Area is obtained by rotating lines shown about the x–axis.
L, in.
, in.y
2
, inyL
2
2 2 (2.3117 in )A yL
ππ
=Σ=
2
14.52 inA=
consent of McGraw–Hill Education.
PROBLEM 5.38
Determine the volume and the surface area of the solid obtained
by rotating the area of Problem 5.4 about (a) the line x = 240 mm,
(b) the y axis.
SOLUTION
From the solution to Problem 5.4 we have
32
63
63
15.3 10 mm
2.6865 10 mm
1.4445 10 mm
A
xA
yA
= ×
Σ= ×
Σ= ×
Applying the theorems of Pappus-Guldinus we have
(a) Rotation about the line
240 mmx=
36
Volume 2 (240 )
2 (240 )
2 [240(15.3 10 ) 2.6865 10 ]
xA
A xA
π
π
π
= −
= −Σ
= ×− ×
63
Volume 6.19 10 mm
= ×
line line
1133445566
Area 2 2 ( )
2( )
XL x L
xL xL xL xL xL
ππ
π
= = Σ
= ++++
Where
16
,,xx
are measured with respect to line
240 mm.x=
Area 2 [(120)(240) (15)(30) (30)(270)
(135)(210) (240)(30)]
π
= ++
++
32
Area 458 10 mm= ×
(b) Rotation about the y axis
area
63
Volume 2 2 ( )
2 (2.6865 10 mm )
X A xA
ππ
π
= = Σ
= ×
63
Volume 16.88 10 mm= ×
line line
1122334455
Area 2 2 ( )
2( )
2 [(120)(240) (240)(300)
(225)(30) (210)(270) (105)(210)]
XL x L
xL xL xL xL xL
ππ
π
π
= = Σ
= ++++
= +
++ +
62
Area 1.171 10 mm= ×
consent of McGraw–Hill Education.
PROBLEM 5.39
Determine the volume and the surface area of the solid obtained by rotating the
area of Problem 5.8 about (a) the x axis, (b) the y axis.
SOLUTION
From the solution to Problem 5.8 we have
2
3
3
1146.57 in.
14147.0 in.
26897 in.
A
xA
yA
=
Σ=
Σ=
Applying the theorems of Pappus-Guldinus we have
(a) Rotation about the x axis:
3
Volume 2 2
2 (26897 in. )
area
Y A yA
ππ
π
= = Σ
=
or
33
Volume 169.0 10 in= ×
2233445566
Area 2
2( )
2( )
2 [(7.5)(15) (30)( 15) (47.5)(5)
(50)(30) (25)(50)]
line
line
YA
yA
yL yL yL yL yL
π
π
π
ππ
=
= Σ
= ++++
= + ×+
++
or
32
Area 28.4 10 in= ×
(b) Rotation about the y axis
3
Volume 2 2
2 (14147.0 in. )
area
X A xA
ππ
π
= = Σ
=
or
33
Volume 88.9 10 in= ×
1122334455
Area 2 2 ( )
2( )
2 15
2 (15)(30) (30)(15) 30 ( 15) (30)(5) (15)(30)
line line
XL x L
xL xL xL xL xL
ππ
π
ππ
π
= = Σ
= ++++
×
= + + − ×+ +
or
32
Area 15.48 10 in= ×
consent of McGraw–Hill Education.
PROBLEM 5.40
Determine the volume of the solid generated by rotating the
parabolic area shown about (a) the x-axis, (b) the axis AA′.
SOLUTION
First, from Figure 5.8a, we have
4
3
2
5
A ah
yh
=
=
(a) Rotation about the x–axis:
Volume 2
24
253
yA
h ah
π
π
=
=
or
2
16
Volume 15 ah
π
=
(b) Rotation about the line
:AA′
Volume 2 (2 )
4
2 (2 ) 3
aA
a ah
π
π
=
=
or
2
16
Volume 3ah
π
=
consent of McGraw–Hill Education.
PROBLEM 5.41
Determine the capacity, in liters, of the punch bowl
shown if R = 250 mm.
SOLUTION
The volume can be generated by rotating the triangle and circular sector shown about the y-axis. Applying
the second theorem of Pappus-Guldinus and using Figure 5.8a, we have
11 2 2
2
6
33
3
3
3
22
2( )
1 1 1 1 3 2 sin30
232 22 2 3 6
2163 23
33
8
33 (0.25 m)
8
0.031883 m
V xA xA
xA xA
R
R RR R
RR
R
π
ππ
π
π
π
π
π
π
= = Σ
= +
°
= × ×× +
×
= +
=
=
=
Since
33
10 l 1 m=
3
33
10 l
0.031883 m 1m
V= ×
31.9 litersV=
consent of McGraw–Hill Education.
PROBLEM 5.42
The aluminum shade for the small high–intensity lamp shown has a uniform thickness of 1 mm. Knowing
that the density of aluminum is 2800 kg/m3, determine the mass of the shade.
SOLUTION
The mass of the lamp shade is given by
m V At
ρρ
= =
where A is the surface area and t is the thickness of the shade. The area can be generated by rotating the line
shown about the x-axis. Applying the first theorem of Pappus Guldinus, we have
11 22 33 44
22
2( )
A yL yL
yL yL yL yL
ππ
π
= = Σ
= +++
or
22
22
22
2
13 mm 13 16
2 (13 mm) mm (32 mm) (3 mm)
22
16 28 mm (8 mm) (12 mm)
2
28 33 mm (28 mm) (5 mm)
2
2 (84.5 466.03 317.29 867.51)
10,903.4 mm
A
π
π
+
= + ×+
+
+ ×+
+
+ ×+
= +++
=
Then
3 32
(2800 kg/m )(10.9034 10 m )(0.001 m)
m At
ρ
−
=
= ×
or
0.0305 kgm=
consent of McGraw–Hill Education.
PROBLEM 5.43
Knowing that two equal caps have been removed from a 10–in.–diameter
wooden sphere, determine the total surface area of the remaining portion.
SOLUTION
The surface area can be generated by rotating the line shown about the y-axis. Applying the first theorem of
Pappus-Guldinus, we have
11 2 2
22
2 (2 )
A XL x L
xL xL
ππ
π
= = Σ
= +
Now
4
tan 3
a
=
or
53.130
a
= °
Then
2
180
5 in. sin53.130
53.130
4.3136 in.
x
π
°
×°
=°×
=
and
2
2 53.130 (5 in.)
180
9.2729 in.
3
2 2 in. (3 in.) (4.3136 in.)(9.2729 in.)
2
L
A
π
π
= °×
°
=
= +
or
2
308 inA=
consent of McGraw–Hill Education.
PROBLEM 5.44
Three different drive belt profiles are to be
studied. If at any given time each belt
makes contact with one–half of the
circumference of its pulley, determine the
contact area between the belt and the
pulley for each design.
SOLUTION
Applying the first theorem of Pappus-Guldinus, the contact area
C
A
of a
belt is given by
C
A yL yL
ππ
= = Σ
where the individual lengths are the lengths of the belt cross section that
are in contact with the pulley.
(a)
11 2 2
[2( ) ]
0.125 0.125 in.
2 3 in. [(3 0.125) in.](0.625 in.)
2 cos20
C
A yL yL
π
π
= +
= − +−
°
or
2
8.10 in
C
A=
(b)
11
[2( )]
0.375 0.375 in.
2 3 0.08 in.
2 cos20
C
A yL
π
π
=
= −−
°
or
2
6.85 in
C
A=
(c)
11
[2( )]
2(0.25)
3 in. [ (0.25 in.)]
C
A yL
π
ππ
π
=
= −
or
2
7.01in
C
A=
consent of McGraw–Hill Education.
PROBLEM 5.45
Determine the volume and weight of the solid brass knob shown, knowing
that the specific weight of brass is 0.306 lb/in3.
SOLUTION
Area, in2
, in.y
3
, inyA
1
2
(0.75) 0.4418
4
π
=
0.8183 0.3615
2
(0.5)(0.75) 0.375=
0.25 0.0938
3
(1.25)(0.75) 0.9375=
0.625 0.5859
4
2
(0.75) 0.4418
4
π
−= −
0.9317 −0.4116
Σ
0.6296
33
2 2 (0.6296 in ) 3.9559 inV yA
ππ
=Σ= =
3
3.96 inV=
33
(0.306 lb/in )(3.9559 in )WV
γ
= =
1.211lbW=
Volume of knob is obtained by rotating area
of components shown below.
consent of McGraw–Hill Education.
PROBLEM 5.46
Determine the total surface area of the solid brass knob shown.
SOLUTION
Area is obtained by rotating lines shown about the x–axis.
L, in.
, in.y
2
, inyL
2
2 2 (2.3117 in )A yL
ππ
=Σ=
2
14.52 inA=
consent of McGraw–Hill Education.