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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.
