9.

3 Theorems of Pappus and Guldinus

9.3 Theorems of Pappus and Guldinus Example 1, page 1 of 2

1. Determine the amount of paint required to paint
the inside and outside surfaces of the cone, if one
gallon of paint covers 300 ft2.
y

1 The y axis is the axis of revolution.

y
3 ft
3 ft
3 ft
= 1.5 ft
2

2 The generating curve is a

straight line through the
origin.

10 ft

10 ft

x
x

3 The distance to the centroid of the line is

rC = 1.5 ft

9.3 Theorems of Pappus and Guldinus Example 1, page 2 of 2

3 ft

4 The length of the curve can be found from the Pythagorean theorem:
L=

(10 ft)2 + (3 ft)2

= 10.4403 ft

10 ft

Applying the first theorem of Pappus-Guldinus gives the area:

A = 2 rcL
= 2 (1.5 ft)(10.4403 ft)
= 98.3975 ft2
Calculate the volume of paint required:
Volume of paint = 2(98.3975 ft2)(

= 0.656 gal

1 gal
300 ft2

Ans.

Because both the inside and outside surfaces must be painted, the value of
the computed area must be doubled.

9.3 Theorems of Pappus and Guldinus Example 2, page 1 of 1

2. Determine the volume of the cone.
y
1 The y axis is the axis of rotation.
y

3 ft
rc =

3 ft
= 1 ft
3

3 ft
2 The generating area is
a triangle

10 ft

3 The centroid of a
triangle is located
one-third of the
distance from the base
to the opposite vertes.

C
10 ft
4 The area of the triangle is
x

A = (1/2)(10 ft)(3 ft)

= 15 ft2

Applying the second theorem of

Pappus-Guldinus gives the volume:
V = 2 rcA
=2

ft)( 15 ft2)

= 94.2 ft3

Ans.

9.3 Theorems of Pappus and Guldinus Example 3, page 1 of 1

3. Determine the area of the
half-torus (half of a doughnut).

1 The axis of revolution is the x axis.

y
1m

2 The generating curve

is a circle.

y
C

4 The length of
the curve is the
circumference
of the circle:

1m

4m

C
x

4m

L = 2 (1m)
x

= 6.2832 m

z
3

The distance to the centroid

is rc = 4 m.
5 Applying the first theorem of
Pappus-Guldinus gives the area:
A = rcL
= (4 m)(6.2832 m)
= 79.0 m2

Ans.

The angle of revolution is , not 2 , because

the figure is a half -torus.

9.3 Theorems of Pappus and Guldinus Example 4, page 1 of 1

4. Determine the volume of the
half-torus (half of a doughnut).

1 The axis of revolution is the x axis

y
1m

2 The generating area is the

region bounded by a circle

y
C

4 The area bounded by

the circle is

1m

4m

C
x

A = (1 m)2

4m

= m2
x

z
3 The distance to the centroid is rc = 4m
5

Applying the second theorem of

Pappus-Guldinus gives the volume:
V = rcA
= (4 m)( m2)
= 39.5 m3

Ans.

The angle of revolution is , not 2 , because

the figure is a half -torus.

9.3 Theorems of Pappus and Guldinus Example 5, page 1 of 2

5. Determine the area of the frustum of the cone.

1 The y axis is the axis of rotation

y

2 The generating curve is the

straight line BD. The
horizontal coordinate d of
the lower end of the line
can be found by similar
triangles:
3m
D

3m

3m

D
4m

4m
d
B

2m
O

2m
x

4m

C
d

2m

O
3m
d
=
2m
2m+4m
Solving gives
d=1m

9.3 Theorems of Pappus and Guldinus Example 5, page 2 of 2

y
3m

The distance to
the centroid is
rc = (1 m + 3 m)/2

C
B

1m

=2m

3m

O
D
4 The length of the generating curve BD is
4m
B
O

3m

L = (2 m)2 + (4 m)2

2m

= 4.4721 m

4m

1m=2m
D

x
B

5 Applying the first theorem of Pappus-Guldinus gives the area:

A = 2 rcL
= 2 (2 m)(4.4721 m)
= 56.2 m2

Ans.

9.3 Theorems of Pappus and Guldinus Example 6, page 1 of 3

6. Determine the volume of the frustum of the cone.
1 The y axis is the axis of rotation
y
rC
3m
2 The generating area is the
trapezoid BDEF. The
horizontal coordinate d of
D
E
the lower end of the line
C
BD can be found by similar
4m
triangles:
3m
B
F
D
d
2m
4m
x
O
d
B
2m

y
3m

4m

2m
O

O
3 The distance rc to the centroid of the
area can be calculated by dividing the
crosshatched trapezoid into composite
parts and using the formula

rc =

xelAel
Ael

3m
d
=
2m
2m+4m
Solving gives
d=1m

(1)

where xel is the centroidal coordinate

of the part with area Ael.

9.3 Theorems of Pappus and Guldinus Example 6, page 2 of 3

4

However, we can save some work by noting that the

second theorem of Pappus-Guldinus involves the
product of rc and the generating area A:
V = 2 rcA

(2)

where V is the volume of the solid of revolution.

Solving Eq. 1 for the product of rc and Ael gives
rc Ael = xelAel
A = total area = sum of individual elements of area
and substituting this result in Eq. 2 gives
V=2

xelAel

(3)

Thus we do not need to calculate rc and A

independently we need only evaluate the first
moment of the area, xelAel.

9.3 Theorems of Pappus and Guldinus Example 6, page 3 of 3

5

To calculate xelAel, divide the trapezoidal region into

the sum of a rectangle and triangle, and set up a table.
y

y
3m

2m

1m

1m
Ael = (4 m)(1 m)

4m

4m

= 4 m2

1m
O

Ael = (1/2)(4 m)(2 m)

C
+

4m

xel = 0.5 m
x

4m

= 4 m2

xel = 1 m + (1/3)(2 m)
Region

Ael (m2)

xel (m)

Rectangle
Triangle

4
4

0.5
1.6667

xelAel (m3)

= 1.6667 m

2
6.6668
xelAel = 8.6668

Substituting the value of xelAel into Eq. 3 gives the volume of the solid:
V=2

xelAel

(Eq. 3 repeated)
8.6668

= 54.5 m3

Ans.

9.3 Theorems of Pappus and Guldinus Example 7, page 1 of 1

7. Determine the centroidal coordinate rc of a
semicircular arc of radius R, given that the area of a
sphere of radius R is known to be 4 R2.
y

1 If the semicircle is revolved around the y axis,

a sphere of radius R is generated. The first
theorem of Pappus-Guldinus says that the
area of the sphere is given by
A = 2 rcL

rC
C

Because we already know A (= 4 R2), we can

solve this equation for rc in terms of R and L:

4 R2
rc =

A
2 L

(1)

2 The length L in Eq. 1 is the circumference of

the semicircle:
L= R
Substituting this result in Eq. 1 gives
2
rc = 4 R
2 L

(Eq. 1. repeated)
R

2R

Ans.

9.3 Theorems of Pappus and Guldinus Example 8, page 1 of 1

8. Determine the centroidal coordinate rc of a
semicircular area of radius R, given that the volume of a
sphere is known to be (4/3) R3.
y
Radius = R

1 If the semicircular area is revolved around the y axis,

a sphere of radius R is generated. The second
theorem of Pappus-Guldinus says that the volume of
the sphere is given by
V = 2 rcA

rC
C

(1)

Because we already know V (= (4/3) R3), we can

solve Eq. 1 for rc in terms of A and R:
(4/3) R3
rc =

V
2 A

(2)

2 The area A in Eq. 2 is the area bounded by the

semicircle:
R2
2
Substituting this result in Eq. 2 gives
A=

rc =

(4/3) R3
2 A

(Eq. 2 repeated)
R2
2

4R
3

Ans.

9.3 Theorems of Pappus and Guldinus Example 9, page 1 of 4

9. A concrete dam is to be constructed in the shape
shown. Determine the volume of concrete that
would be required.
1 The y axis is the axis of revolution
y

3.5 m
20 m

3.5 m

x
40

20 m
3m

3m

1m 2m

1m 2m

2 The crosshatched region is the generating area.

9.3 Theorems of Pappus and Guldinus Example 9, page 2 of 4

y
rc

x
20 m
1m 2m

Generating area divided into a

rectangle and two triangles

The second theorem of Pappus-Guildinus gives the

volume as
V = 2 40/360)rcA

3.5 m

3m

(1)

where rc is the distance to the centroid of the generating

area, A is the magnitude of the area, and the
factor 40/360 accounts for the fact that the dam
corresponds to 40 rather than to a complete circle.
Thus we must calculate the product rcA. This product
may be found by dividing the crosshatched area into
composite parts and then using the formula

rc =

xelAel
Ael

(2)
A = sum of element areas

where xel is the centroidal coordinate of the part with

area Ael. Solving for the product rcA gives
rcA = xelAel
Thus Eq. 1 can be written as
V = 2 40/360)(rcA)
= (2/9)

xelAel

(Eq. 1 repeated)
(3)

and, to calculate V, we have to evaluate the sum xelAel.

9.3 Theorems of Pappus and Guldinus Example 9, page 3 of 4

y

4 Calculate the areas and centroidal

distances, and set up a table.

xel = 23 m + (3 m)/3 = 24 m

3.5 m
Ael = (1/2)(3.0 m)

x
23 m

3m

rc

Triangle 1

+
xel = 20 m + 2 m + 0.5 m = 22.5 m

Ael = (3.5 m)

3.5 m

20 m
1m 2m

3.5 m

3m

3.5 m) = 5.25 m2

m) = 3.5 m2
O

x
Rectangle
1m 2m

20 m
O
+

xel = 20 m + (2 m)(2/3) = 21.333 m

Ael = (1/2)(3.5 m)
= 3.5 m2

3.5 m
x

m)
O

Triangle 2

2m

20 m

9.3 Theorems of Pappus and Guldinus Example 9, page 4 of 4

5 Table
Region

Ael (m2)

xel (m)

Triangle 1
Rectangle
Triangle 2

5.25
3.5
3.5

24
22.500
21.333

xelAel (m3)
126
78.750
74.665
xelAel = 279.415

6 Substituting the value of xelAel into Eq. 3 gives the

volume of the solid:
2
V= 9

xelAel

(Eq. 3 repeated)

279.415 m3
= 195.1 m3

Ans

9.3 Theorems of Pappus and Guldinus Example 10, page 1 of 4

10. The concrete steps shown are in the shape of a quarter
circle. Determine the amount of paint required to paint
the steps, if one liter of paint covers 1.5 m2.
y
260 mm

190 mm

260 mm

190 mm
1 The y axis is the axis of revolution.

x
z

y
2 The generating
curve is a series of
four straight
line-segments.

190 mm
190 mm

x
260 mm

260 mm

9.3 Theorems of Pappus and Guldinus Example 10, page 2 of 4

3 The first theorem of Pappus-Guildinus gives the area as
A = 2 (90/360)rcL
y

= ( /2)rcL

190 mm
190 mm
x
260 mm

(1)

where rc is the distance to the centroid of the generating

curve, L is the length of the area, and the factor (90/360)
accounts for the fact that the steps are in the shape of a
quarter circle. Thus we must calculate the product rcL.
This product may be found from the formula for the centroid
of a composite curve made of a of collection of line
segments:

260 mm
rc =

xelLel
Lel

(2)
L = sum of segment lengths

where xel is the centroidal coordinate of the line segment

with length Lel. Solving for the product rcL gives
rcL = xelLel
Thus Eq. 1 can be written as
A = ( /2)(rcL)
= ( /2) xelLel

(Eq. 1 repeated)
(3)

and, to calculate A, we have to evaluate the sum xelLel.

9.3 Theorems of Pappus and Guldinus Example 10, page 3 of 4

4

Calculate the centroidal coordinates

and lengths of the line segments, and
set up a table.

Lel = 190 mm
Lel = 260 mm

y
y

xel = 260 mm

260 mm
=

190 mm

190 mm

xel = 260 mm/2

= 130 mm

380 mm

+
190 mm
x

190 mm
x
260 mm

Line 2

Line 1

260 mm
y

xel = 260 mm + 260 mm

= 520 mm

Lel = 260 mm
+

260 mm

Lel = 190 mm

260 mm
+
190 mm
x

190 mm
x
260 mm

260 mm

xel = 260 mm + (260 mm)/2 = 390 mm

Line 3

Line 4

9.3 Theorems of Pappus and Guldinus Example 10, page 4 of 4

5 Table
Region

Lel (mm)

xel (mm)

Line 1
Line 2
Line 3
Line 4

260
190
260
190

130
260
390
520

xelLel (mm2)
33 800
49 400
104 100
98 800
xelLel = 283 400

6 Substituting the value of xelAelinto Eq. 3 gives the area of the steps
A = 2 xelLel

(Eq. 3 repeated)
2

283 400 mm
= 445 164 mm2
7

Amount of paint required = 445 164 mm2

= 0.297 liter

[1 m/(1000 mm)]2

[1 liter of paint/1.5 m2 of covered area]

Ans.

9.3 Theorems of Pappus and Guldinus Example 11, page 1 of 3

11. The concrete steps shown are in the shape of a quarter circle.
Determine the total number of cubic meters of concrete required to
construct the steps.
y

The y axis is the axis of revolution

260 mm

190 mm

190 mm
260 mm
190 mm
x

190 mm

260 mm

260 mm

x
z

2 The generating area is the crosshatched region

shown.

9.3 Theorems of Pappus and Guldinus Example 11, page 2 of 3

3 The second theorem of Pappus-Guildinus gives the
volume as
V = 2(90/360 rcA
= ( /2)rcA

190 mm
190 mm
O

260 mm

x
260 mm

Generating area divided into two

rectangles

(1)

where rc is the distance to the centroid of the generating

area, A is the magnitude of the area, and the factor
(90/360) accounts for the fact that the steps form a
quarter circle. Thus we must calculate the product rcA.
This product may be found by dividing the crosshatched
area into composite parts and then using the formula
rc =

xelAel
Ael

(2)
A = sum of element areas

where xel is the centroidal coordinate of the part with

area Ael. Solving for the product rcA gives
rcA = xelAel
Thus Eq. 1 can be written as
V = ( /2)(rcA)
= ( /2) xelAel

(Eq. 1 repeated)
(3)

and, to calculate V, we have to evaluate the sum xelAel.

9.3 Theorems of Pappus and Guldinus Example 11, page 3 of 3

4

Calculate the centroidal distances and areas

and set up a table.
y

190 mm

260 mm

xel = 260 mm + 260 mm/2 = 390 mm

380 mm

190 mm
O

xel = 260 mm/2 = 130 mm y

O 260 mm

260 mm
Ael = (380 mm)(260 mm)
= 98 800 mm2

Region

Ael (mm2)

xel (mm)

Rectangle 1
Rectangle 2

98 800
49 400

130
390

xelAel (mm3)

12 844 000
19 266 000
xelAel = 32 110 000

Substituting the value of xelAelinto Eq. 3 gives the volume of the solid
V = 2 xelAel

32 110 000 mm3

= 50 438 270 mm3

= 0.0504 m3

O
260 mm

Rectangle 1

190 mm
x

Ans.

260 mm

Rectangle 2
Ael = (190 mm)(260 mm)
= 49 400 mm2

9.3 Theorems of Pappus and Guldinus Example 12, page 1 of 4

12. Determine the mass of the steel V-belt pulley
shown. The density of the steel is 7840 kg/m3.

12 mm 10 mm 12 mm
4 mm

4 mm

The x axis is the axis

of revolution:

15 mm

(20 mm)/2 = 10 mm

100 mm 70 mm
2
20 mm
15 mm

Front view

Side view

The crosshatched area is

the generating area for
the right half of the
pulley.

9.3 Theorems of Pappus and Guldinus Example 12, page 2 of 4

3 The second theorem of Pappus-Guildinus gives the
volume as
y

5 mm 12 mm

V = 2 rcA

(1)

4 mm
15 mm
(70 mm
10 mm

20 mm)/2 = 25 mm
x

Generating area divided into two

rectangles, and a triangle

where rc is the distance to the centroid of the generating

area, and A is the magnitude of the area. Thus we must
calculate the product rcA. This product may be found
by dividing the crosshatched area into composite parts
and then using the formula

rc =

yelAel
Ael

(2)
A = sum of element areas

where xel is the centroidal coordinate of the part with

area Ael. Solving for the product rcA gives
rcA = yelAel
Thus Eq. 1 can be written as
V = 2 (rcA)
=2

yelAel

(Eq. 1 repeated)
(3)

and, to calculate V, we have to evaluate the sum yelAel.

9.3 Theorems of Pappus and Guldinus Example 12, page 3 of 4

4 Calculate the centroidal distances
and areas for the rectangles and
triangle, and set up a table.
y

5 mm 12 mm

Ael = (21 mm)(25 mm)

= 525 mm2
y

4 mm

Ael = (1/2)(12 mm)(15 mm)

= 90 mm2
y 12 mm

5 mm + 12 mm + 4 mm
= 21 mm

15 mm

15 mm
+
25 mm
10 mm

=
x

Generating area divided into

two rectangles, and a triangle

25 mm
10 mm

10 mm + 5 mm = 35 mm
x

Rectangle 1
yel = 10 mm + 25/2 mm
= 22.5 mm
yel = 35 mm + 15/3 mm
= 40 mm
+
yel = 35 mm + 15/2 mm
= 42.5 mm
Ael = (4 mm)(15 mm)
= 60 mm2

Triangle
y
4 mm
15 mm

35 mm
x
Rectangle 2

9.3 Theorems of Pappus and Guldinus Example 12, page 4 of 4

5

Table

Region

Ael (mm2)

yel (mm)

Rectangle 1
Triangle
Rectangle 2

525
90
60

22.5
40.0
42.5

yelAel (mm3)

11 812.5
3 600.0
2 550.0
yelAel = 17 962.5

Substituting the value of yelAelinto Eq. 3 gives half the volume of the
V-belt pulley
V=

yelAel
17 962.5 mm3

= 112 861.7 mm3 = 0.000 112 861 7 m3

7

Total mass of the V-belt = total volume

= (2

0.000 112 861 7 m3)

= 1.770 kg
Double the half-volume.

density
7840 kg/m3
Ans.

9.3 Theorems of Pappus and Guldinus Example 13, page 1 of 3

13. Determine the area of the surface of
revolution generated by rotating the curve
y = z4, 0 z 1 m, about the z axis.
y
1 The z axis is the axis of revolution.
2 The generating curve is

y = z4, 0 z

y
1m
1m
z
z
1m

9.3 Theorems of Pappus and Guldinus Example 13, page 2 of 3

3

The first theorem of Pappus-Guildinus gives the area of

the surface of revolution as
A = 2 rcL

(1)

where rc is the distance to the centroid of the generating

curve, and L is the length of the curve. Thus we must
calculate the product rcL. This product may be found
by considering the equation for the centroidal
coordinate:

rc =

1m

rc
z
1m

yel dL
dL

L
y

where dL is an increment of curve length, and yel is the

coordinate of the increment. Solving for the product rcL
gives

dL
1m

rcL = yel dL

yel

Thus Eq. 1 can be written as

z
1m

A = 2 (rcL)
=2

yel dL

(2)

and, to calculate A, we have to evaluate a single integral,

yeldL.

9.3 Theorems of Pappus and Guldinus Example 13, page 3 of 3

4

To evaluate the integral yel dL in Eq. 2, we use the

equation of the curve,
y = z4

(3)

to express dL as a function of z. Thus

dL =
=

(dy)2 + (dz)2
( dy )2 + 1 dz
dz

(4)

y
dL

and differentiating Eq. 3 gives

dz

dy = 4z3
dz
so Eq. 4 can be written as
dL =

yel = y = z4

(4z3)2 + 1 dz

(5)
1m

Substitute this expression for dL into Eq. 2

A=2

dy dL

yel dL

1
= 2 z4 (4z3)2 + 1 dz
0

(Eq. 2 repeated)

5 Evaluating the integral by use of the integral

function key on a calculator gives
A = 3.44 m2

Ans.

9.3 Theorems of Pappus and Guldinus Example 14, page 1 of 3

14. Determine the volume of the solid of
revolution generated by rotating the curve
y = z4, 0 z 1 m, about the z axis.
y

The z axis is the axis of rotation.

2 The generating area is the

area under the y = z4 curve.
y
y = z4
x
1m

1m

z
1m

9.3 Theorems of Pappus and Guldinus Example 14, page 2 of 3

y

The second theorem of Pappus-Guildinus gives the

volume as
V = 2 rcA

y = z4
C

rC

where rc is the distance to the centroid of the generating

area, and A is the magnitude of the area. Thus we must
calculate the product rcA. This product may be found
by considering the equation for the centroidal distance:

rc =

y = z4

(1)

yel dA
dA

where dA is an increment of area, and yel is the

coordinate of the centroid of the incremental region.
Solving for the product rca gives

(x, y)
yel

rcA = yel dA
Thus Eq. 1 can be written as
V = 2 (rcA)
=2

yel dA

(2)

and, to calculate V, we have to evaluate a single integral,

yel dA.

9.3 Theorems of Pappus and Guldinus Example 14, page 3 of 3

4 To evaluate the integral yel dL in Eq. 2, we use the
equation of the curve,
y = z4
to express dA as a function of z. Thus
y

dA = y dz
= z4 dz
y = z4

(y, z)

yel = y/2 = z4/2

Substitute this expression for dA into Eq. 2

A=2

yel dz

(Eq. 2 repeated)

z
dz
1m

=2

z4/2)(z4) dz
(
0

Evaluating the integral by use of the integral function key

on a calculator gives
V = 0.349 m2

Ans.

9.3 Theorems of Pappus and Guldinus Example 15, page 1 of 5

15. A pharmaceutical company plans to put a coating
0.01 mm thick on the outside of the pill shown.
Determine the amount of coating material required.
Radius = 20 mm

1 The x axis is the axis of revolution.

y
(1.5 mm)/2 = 0.75 mm

3.5 mm
x
7 mm

1.5 mm
2 The generating curve for half of the
pill surface is a composite curve
consisting of one straight line and a
circular arc. By symmetry, the total
surface area of the pill will be two
times the area generated by the curve
above.

9.3 Theorems of Pappus and Guldinus Example 15, page 2 of 5

3 The first theorem of Pappus-Guildinus gives the area as
A = 2 rcL
y
0.75 mm

where rc is the distance to the centroid of the generating

curve and L is the length of the curve. Thus we must
calculate the product rcL. This product may be found
by dividing the curve into composite parts and then
using the formula

3.5 mm
rc =
x
Generating curve divided into a
straight line segment and an arc

(1)

yelLel
Lel

(2)
L

where xel is the centroidal coordinate of the part with

length Lel. Solving for the product rcL gives
rcL = yelLel
Thus Eq. 1 can be written as
A = 2 (rcL)
=2

yelLel

(Eq. 1 repeated)
(3)

and, to calculate A, we have to evaluate the sum yelLel.

9.3 Theorems of Pappus and Guldinus Example 15, page 3 of 5

4 Calculate the centroidal distances and
the lengths, and set up a table.
y

y
0.75 mm

0.75 mm

yel = 3.5 mm

3.5 mm

=
x
Radius = 20 mm

3.5 mm

x
20 mm

Straight line

Arc
y
5 For the straight line, the length and
coordinate of the centroid are easily
calculated..
Lel = 0.75 mm

0.75 mm

yel = 3.5 mm
x
Straight line

9.3 Theorems of Pappus and Guldinus Example 15, page 4 of 5

6 To calculate xelLel for the arc,
use the information shown below,
which has been taken from a table
of properties of common
geometric shapes.

A
20 mm

3.5 mm

7 For our particular arc, r = 20 mm and

= (1/2) sin-1(3.5/20)

2
B

= 0.08795 rad
Thus

Centroid Location
Length = 2 r

y
r

Length = 2 r
C

= 2(0.08795)

20 mm

= 3.5180 mm
rarc = (r sin
= (20 mm)(sin 0.08795)/(0.08795)

r sin

= 19.9742 mm

Circular arc segment

y

yel = rarc sin

A

19.9742 mm (sin 0.08795)

rarc

= 1.7545 mm

3.5 mm
yel
B

20 mm
Arc

9.3 Theorems of Pappus and Guldinus Example 15, page 5 of 5

8 Table
yelLel (m2)

Region

Lel (m)

yel (m)

Line
Arc

0.75
3.5180

3.5
2.6250
1.7545
6.1723
yelLel = 8.7973

Substituting the value of yelAelinto Eq. 3 gives the area of the solid
A=2

yelLel
8.7973 mm2

= 110.5501 mm2
where a factor of 2 has been inserted to account for the fact that we took advantage
of symmetry to calculate the area of only half of the body.
Amount of coating material required = 110.5501 mm2
= 1.106 mm3

0.01 mm
Ans.

9.3 Theorems of Pappus and Guldinus Example 16, page 1 of 4

16. Determine the volume of the funnel.
1

The y axis is the axis of revolution

10 mm

y
10 mm
60 mm

5 mm

60 mm
5 mm
70 mm

5 mm

70 mm

2.5 mm

2 The generating area is the

crosshatched area shown.

9.3 Theorems of Pappus and Guldinus Example 16, page 2 of 4

3 The second theorem of Pappus-Guildinus gives the
volume as

y
10 mm

V = 2 rcA

5 mm

where rc is the distance to the centroid of the generating

area and A is the magnitude of the area. Thus we must
calculate the product rcA. This product may be found
by dividing the cross-hatched area into composite parts
and then using the formula

60 mm

rc =

70 mm

2.5 mm

(1)

Generating area divided into two

rectangles and two triangles

xelAel
Ael

(2)
A

where xel is the centroidal coordinate of the part with

area Ael. Solving for the product rcA gives
rcA = xelAel
Thus Eq. 1 can be written as
V = 2 (rcA)
=2

xelAel

(Eq. 1 repeated)
(3)

and, to calculate V, we have to evaluate the sum xelAel.

9.3 Theorems of Pappus and Guldinus Example 16, page 3 of 4

4

To calculate xelAel, divide the area

into the sum of two rectangles and
two triangles, and set up a table.
y

10 mm

5 mm
xel = 2.5 mm

y 10 mm
xel = 5 mm + (5 mm)/3
= 6.6667 mm
5 mm
60 mm

60 mm

5 mm

Ael = (5 mm)(60 mm)

= 300 mm2

60 mm
Ael = (1/2)(5 mm)(60 mm)
= 150 mm2
O

=
70 mm

2.5 mm

Rectangle 1

x
Triangle 1

y
y

2.5 mm

2.5 mm

xel =1.25 mm
xel = 2.5 mm + (2.5 mm)/3
= 3.3333 mm
+

+
70 mm
Ael = (2.5 mm)(70 mm)
= 175 mm2
x
Rectangle 1

70 mm
Ael = (1/2)(2.5 mm)(70 mm)
= 87.5 mm2
x
2.5 mm
Triangle 2

9.3 Theorems of Pappus and Guldinus Example 16, page 4 of 4

5 Table
Region

Ael (mm2)

xel (mm)

Rectangle 1
Rectangle 1
Triangle 1
Triangle 2

300
175
150
87.5

2.5
750
1.25
218.75
6.6667
1000.0050
3.3333
291.6637
xelAel = 2260.4187

xelAel (mm3)

Substituting the value of xelAelinto Eq. 3 gives the volume of the funnel
V=

xelAel

(Eq. 3 repeated)

2260.4187 mm3
= 14 200 mm3

Ans.

9.3 Theorems of Pappus and Guldinus Example 17, page 1 of 3

17. A satellite dish is shaped in the form of a paraboloid of
revolution to take advantage of the geometrical fact that all
signals traveling parallel to the axis of the paraboloid are
reflected through the focus. Determine the amount, in m2,
of reflecting material required to cover the inside surface of
the dish.

The x axis is the axis of revolution

y
(0, 0.3)

0.3 m

( 0.2, 0)
Signals parallel
to axis of dish

2 The generating
curve is a parabola.

3 The general form for a parabola with vertex on the x

axis is
x = ay2 + b

0.3 m

Evaluating this equation at the points ( 0.2, 0) and

(0, 0.3) gives the equations
0.2 = a(0)2 + b
0.2 m

and

(1)

0 = a(0.3)2 + b

Solving for a and b and substituting back in Eq. 1

gives
x = 0.2222y2

0.2

(2)

9.3 Theorems of Pappus and Guldinus Example 17, page 2 of 3

4 Apply the first theorem of Pappus-Guldinus
to calculate the surface area of the dish:
A = 2 rcL

y
dL

yel
rc
x

( 0.2, 0)
5

yel dL
dL

(0, 0.3)

dx

(3)

Thus we must calculate the product of the

length of the parabolic curve and its
centroidal coordinate. The product may be
found by considering the equation for the
centroidal coordinate:

rc =

dL

dy

To evaluate the integral yel dL in Eq. 4, we have to use

the equation of the parabola,
x = 0.2222y2

0.2

(Eq. 2 repeated)

Solving for the product rcL gives

to express dL as a function of y. Thus
rcL = yel dL

dL =

Thus Eq. 3 can be written as

=
A = 2 (rcL)

(dx)2 + (dy)2
dx
dy

+ 1 dy

(5)

Differentiating Eq. 2 gives

=2

yel dL

(4)

dx = 0.4444y,
dy
so Eq. 5 can be written as
dL =

(0.4444y)2 + 1 dy

(6)

9.3 Theorems of Pappus and Guldinus Example 17, page 3 of 3

6

Noting that y = yel and also using Eq. 6 to replace dL in

Eq. 4 gives

A=2

=2

y
yel dL

y
(0, 0.3)

(Eq. 4 repeated)

(x, y)

0.3
y (0.4444y)2 + 1 dy
0

Using the integral function on a calculator gives

A = 0.284 m2

Ans

yel
rc

( 0.2, 0)

9.3 Theorems of Pappus and Guldinus Example 18, page 1 of 6

18. Determine the amount of coffee that
the coffee mug holds when full to the
brim. The radius of the rounded corners
and the rim is 15 mm.

1 The y axis is the axis of revolution, and the

generating area is the cross-hatched area shown.
y

80 mm
Radius = 15 mm

90 mm

x
Radius = 15 mm

2 The distance rc to the centroid of the area can

be calculated by dividing the crosshatched
area into composite parts and using the
formula

rc =

xelAel
Ael

(1)

where xel is the centroidal coordinate of the

part with area Ael.

9.3 Theorems of Pappus and Guldinus Example 18, page 2 of 6

3

However, we can save some work by noting that the second

theorem of Pappus-Guldinus involves the product of rc and the
generating area A:
V = 2 rcA

(2)

where V is the volume of solid of revolution.

Solving Eq. 1 for the product gives
rc Ael = xelAel
A
and substituting this result in Eq. 2 gives
V=2

xelAel

(3)

Thus we do not need to calculate rc and A independently

need only evaluate the first moment of the area, xelAel.

we

9.3 Theorems of Pappus and Guldinus Example 18, page 3 of 6

4

To calculate xelAel, divide the area into the algebraic

sum of a rectangle, two squares, and two quarter circles,
and set up a table.
y
y
40 mm

For clarity, the areas near the rim and

rounded corner have been drawn
disproportionately large.
y

Radius = 15 mm

90 mm

Radius = 15 mm
5 Subtract quartercircular area
from square to
form area near
rim.

Square 1
6 Subtract square
from quartercircular area to
form rounded
corner.

x
Quarter circle 1

Rectangle

Square 2

x
Quarter circle 2

9.3 Theorems of Pappus and Guldinus Example 18, page 4 of 6

7

Calculate the areas and centroidal coordinates of the

rectangle and the squares.
y
y
40 mm
40 mm
Radius = 15 mm

90 mm

40 mm

15 mm

Ael = (40 mm)(90 mm)

= 3600 mm2

15 mm

90 mm

xel = 40 mm + (15 mm)/2

= 47.5 mm
x

x
xel = 20 mm

Radius = 15 mm
Original area

Square 1

Rectangle
y

40 mm

Ael = (15 mm)2 = 225 mm2

15 mm

xel = 40 mm (15 mm)/2

= 32.5 mm

15 mm
x
Square 2

9.3 Theorems of Pappus and Guldinus Example 18, page 5 of 6

8

Calculate the area and centroidal coordinates

of the quarter-circular regions..

9 A table of properties of planar regions

gives the information shown below.

y
Centroid location
y

40 mm

15 mm

xel = 40 mm + 15 mm
= 48.6338 mm
x
Quarter circle 1

A=

r2
4

6.3662 mm
C
4r
3

r
x

Quarter circular region

y
40 mm
15 mm
xel = 40 mm 15 mm
+ 6.3662 mm
= 31.3662 mm
x
Quarter circle 2

10 In our particular problem, r = 15 mm, so the

distance to the centroid is
4r = 4(15 mm) = 6.3662 mm4
3
3
Also, the area is
(15 mm)2
Ael =
= 176.7146 mm2
4

9.3 Theorems of Pappus and Guldinus Example 18, page 6 of 6

11 Table
Region

Ael (mm2)

xel (mm)

xelAel (mm3)

Rectangle
Square 1
Quarter circle 1
Square 2
Quarter circle 2

3600
225
176.7146
225
176.7146

20
72 000
47.5
10 687.5
48.6338
8 594.3025
32.5
7 312.5
31.3662
5 542.8655
xelAel = 72 323.5630

12 Substituting the value of xelAelinto Eq. 3 gives the volume of the solid
V=

xelAel
72 323.563 0 mm3

= 454 000 mm3

Ans.

9.3 Theorems of Pappus and Guldinus Example 19, page 1 of 6

19. Determine the capacity of the small bottle of lotion if the
bottle is filled half way up the neck.
5 mm

1 The y axis is the axis of revolution.

Half of the
neck is filled:
(5 mm)/2

2.5 mm
17.5 mm
17.5 mm
Radius = 20 mm
Radius = 20 mm

15 mm
15 mm

2 The generating area is the

crosshatched area shown.

9.3 Theorems of Pappus and Guldinus Example 19, page 2 of 6

y

3 The second theorem of Pappus-Guildinus gives the

volume as

2.5 mm

V = 2 rcA

17.5 mm

Radius = 20 mm

(1)

where rc is the distance to the centroid of the generating

area, and A is the magnitude of the area. Thus we must
calculate the product rcA. This product may be found
by dividing the cross-hatched area into composite parts
and then using the formula

15 mm
x
Generating area divided into a rectangle,
two triangles and a circular sector

rc =

xelAel
Ael

(2)
A

where xel is the centroidal coordinate of the part with

area Ael. Solving for the product rcA gives
rcA = xelAel
Thus Eq. 1 can be written as
V = 2 (rcA)
=2

xelAel

(Eq. 1 repeated)
(3)

and, to calculate V, we have to evaluate the sum xelAel.

9.3 Theorems of Pappus and Guldinus Example 19, page 3 of 6

4

Before the areas and centroidal coordinates can be

found, we first must find the distances and angles
shown below.
C

D
17.5 mm

2.5 mm

C
D

20 mm

17.5 mm
O

Radius = 20 mm
2

DC = (20 mm)

(17.5 mm)

15 mm

= 9.6825 mm
DOC = cos-1

17.5
20

= 28.9550

O
(4)

15 mm

20 mm

EB = (20 mm)2
= 13.2288 mm

(15 mm)2

EOB = cos-1

15
20

= 41.4096

(5)

9.3 Theorems of Pappus and Guldinus Example 19, page 4 of 6

5 The areas and centroidal coordinates of
the rectangle and triangles can now be
calculated.

9.6825 mm
C

D
17.5 mm

2.5 mm

17.5 mm

Ael = (1/2)(17.5 mm)(9.6825 mm)

= 84.7219 mm2

C
D

Ael = (1/2)(15 mm)(13.2288 mm)

= 99.2160 mm2

xel = 9.6825 mm/3

= 3.2275 mm

xel = (13.2288 mm)/3

= 4.4096 mm

15 mm

B
13.2288 mm

x
15 mm
E

Triangle 2

Triangle 1

Radius = 20 mm

9.6825 mm
2.5 mm

C
D

C
xel = 9.6825 mm/2
= 4.8412 mm

Ael = (2.5 mm)(9.6825 mm)

= 24.2062 mm2

x
Rectangle

+
O

B
Circular sector

9.3 Theorems of Pappus and Guldinus Example 19, page 5 of 6

6 To calculate the area and centroidal
coordinate of the circular sector, we can use
the information shown below, which has been
taken from a table of properties of planar
regions. Note that in the formula equals
half the angle of the arc.
y

= (180

DOC

EOB)/2

by Eq. 4

by Eq. 5

DOC = 28.9550
= (180

rc

y
D C

Centroid Location

28.9550

41.4096)/2

= 54.8177
r = 20 mm

A= r
r

EOB

90

= 54.8177 + 41.4096
C

= 6.2273
rc =

2r sin

B
xel

Circular sector region

90

x
=

2r sin
3
2(20 mm) sin 54.8177
3(54.8177
/180)

EOB = 41.4096
= 11.3903 mm
xel = rc cos

8 Ael = r2
=(

54.8177/180)

= 382.6997 mm2

20 mm)2

= (11.3903 mm) cos 6.2273

= 11.3231 mm

9.3 Theorems of Pappus and Guldinus Example 19, page 6 of 6

9

Table
Region

Ael (mm2)

xel (mm)

xelAel (mm3)

Triangle 1
Triangle 2
Rectangle
Circular sector

84.7219
99.2160
24.2062
382.6997

3.2275
273.4399
4.4096
437.5029
4.8412
117.1871
11.3231
4333.3470
xelAel = 5161.4768

10 Substituting the value of xelAel into Eq. 3 gives the capacity of the bottle:
V=

xelAel
5161.4768 mm3

= 32 400 mm3

Ans.