© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

10–1.
y

Determine the moment of inertia about the x axis.

y ϭ bn xn
a

b

x
a

Solution
Differential Element. Here x =

a

1

y n . The area of the differential element parallel
b
a 1
to the x axis shown shaded in Fig. a is dA = (a - x)dy = aa - 1 y n bdy.
bn
1
n

Moment of Inertia. Perform the integration,






Ix =

LA

y2dA =
=

L0

b

L0

b

y2aa aay2 -

a
1
n

b


a
1
n

b

1

yn bdy
1

yn + 2 bdy

a
a
n
3n + 1
= c y3 - a 1 ba
by
d3
3
3n
+
1
n
n
b




=



=

b

0

1 3
n
ab - a
bab3
3
3n + 1
ab3
3(3n + 1)

Ans.



Ans:
Ix =
1010

ab3
3(3n + 1)



© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

10–2.
y

Determine the moment of inertia about the y axis.

y ϭ bn xn
a

b

x
a

Solution
Differential Element. The area of the differential element parallel to the y axis
b
shown shaded in Fig. a is dA = ydx = n xndx.
a
Moment of Inertia. Perform the integration,


Iy =

LA

x2dA =


L0

a

x2 a

b n
x dxb
an

a



=



=



=

b n+2
dx
nx
L0 a
a

b
1
b(xn + 3) `
na
a n + 3
0

a3b
n + 3

Ans.



Ans:
Iy =
1011

a3b
n + 3


© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

10–3.
y

Determine the moment of inertia for the shaded area about
the x axis.


100 mm

200 mm

y ϭ 1 x2
50

Solution
1

Differential Element. Here x = 250y2 . The area of the differential
element parallel
1
to the x axis shown shaded in Fig. a is dA = 2x dy = 2250y2dy.

x

Moment of Inertia. Perform the integration,






Ix =

LA

y2dA =


L0

200 mm

= 2250

1

y2 c 2250y2dy d

L0

200 mm

2 7
= 2250 a y2 b 3
7

5

y2dy
200 mm

0



= 457.14(106) mm4




= 457(106) mm4

Ans.



Ans:
Ix = 457 ( 106 ) mm4
1012


© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

*10–4.
y

Determine the moment of inertia for the shaded area about
the y axis.

100 mm

200 mm

y ϭ 1 x2
50

Solution

1

Differential Element. Here x = 250 y2 . The moment of inertia of the differential
element parallel to x axis shown in Fig. a about y axis is

x

1
1
2
2
100250 3
(dy)(2x)3 = x3dy = ( 250y2 ) 3dy =
y2dy.
12
3
3
3
Moment of Inertia. Perform the integration,

dIy =





Iy =

L


dIy =

L0

200 mm

100150 3
y2dy
3

100250 2 5 3
a y2 b
=
3
5



= 53.33(106) mm4



= 53.3(106) mm4

200 mm

0

Ans.




Ans:
Iy = 53.3(106) mm4
1013


© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

10–5.
y

Determine the moment of inertia for the shaded area about
the x axis.

y ϭ x1/ 2

1m

x
1m

Solution
Differential Element. The area of the differential element parallel to the y axis
shown shaded in Fig. a is dA = ydx. The moment of inertia of this element about
the x axis is

dIx = dIx′ + dA ∼
y2



=



=



=

y 2
1
(dx)y3 + ydx a b
12
2
1 3
y dx
3

1 1 3
(x2) dx
3

1 3
x2 dx
3
Moment of Inertia. Perform the integration.




=



Ix =





dIx =
L
L0

1m

2 53
x2
=
15
=

1 3
x2dx
3
1m

0


2 4
m = 0.133 m4
15

Ans.

Ans:
Ix = 0.133 m4
1014


© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

10–6.
y

Determine the moment of inertia for the shaded area about
the y axis.

y ϭ x1/ 2

1m

x
1m

Solution
Differential Element. The area of the 1differential element parallel to the y axis

shown shaded in Fig. a is dA = ydx = x2dx.
Moment of Inertia. Perform the integration,






Iy =

LA

x2dA =

L0

1m

2 7
= x2 3
7
=

1

x2 ( x2dx )

1m

0


2 4
m = 0.286 m4
7

Ans.

Ans:
Iy = 0.286 m4
1015


© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

10–7.
y

Determine the moment of inertia for the shaded area about
the x axis.

y2 ϭ 1 Ϫ 0.5x
1m

x
2m

Solution
Differential Element. Here x = 2(1 - y2). The area of the differential element
parallel to the x axis shown shaded in Fig. a is dA = xdy = 2(1 - y2)dy.

Moment of Inertia. Perform the integration,






Ix =

LA

y2dA =

L0

= 2

1m

L0

y2[2(1 - y2)dy]
1m

(y2 - y4)dy

y5
y3
= 2a - b 3
3

5

1m

0

4 4
=
m = 0.267 m4
15

Ans.



Ans:
Ix = 0.267 m4
1016


© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

*10–8.
y

Determine the moment of inertia for the shaded area about
the y axis.

y2 ϭ 1 Ϫ 0.5x

1m

x
2m

Solution
Differential Element. Here x = 2(1 - y2). The moment of inertia of the differential
element parallel to the x axis shown shaded in Fig. a about the y axis is


∼2
dIy = dIy' + dAx



=



=



=

1
x 2
(dy)x3 + xdya b
12
2

1 3
x dy
3

1
32(1 - y2) 4 3 dy
3

8
( - y6 + 3y4 - 3y2 + 1)dy
3
Moment of Inertia. Perform the integration,




=

Iy =

L

dIy =



=




=

8
3 L0

1m

( -y6 + 3y4 - 3y2 + 1)dy

1m
y7
8
3
a - + y5 - y3 + yb `
3
7
5
0

128 4
m = 1.22 m4
105

Ans.



Ans:
Iy = 1.22 m4
1017



© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

10–9.
Determine the moment of inertia of the area about the x
axis. Solve the problem in two ways, using rectangular
differential elements: (a) having a thickness dx and
(b) having a thickness of dy.

y
y = 2.5 – 0.1x2
2.5 ft
5 ft

x

SOLUTION
(a) Differential Element: The area of the differential element parallel to y axis is
dA = ydx. The moment of inertia of this element about x axis is
'
dIx = dIx¿ + dAy 2
=

y 2
1
1dx2y 3 + ydxa b
12
2


=

1
12.5 - 0.1x223 dx
3

=

1
1 - 0.001x6 + 0.075x4 - 1.875x2 + 15.6252 dx
3

Moment of Inertia: Performing the integration, we have
5 ft

Ix =

L

dIx =

1
1 -0.001x6 + 0.075x4 - 1.875x2 + 15.6252 dx
3 L-5 ft

=

5 ft
1

0.075 5
1.875 3
0.001 7
ax +
x x + 15.625xb `
3
7
5
3
-5 ft

= 23.8 ft4

Ans.

(b) Differential Element: Here, x = 225 - 10y. The area of the differential
element parallel to x axis is dA = 2xdy = 2 225 - 10y dy.
Moment of Inertia: Applying Eq. 10–1 and performing the integration, we have
Ix =

LA

y2dA
2.5 ft

= 2

y2 225 - 10ydy

L0


= 2c -

2.5 ft
y2
2y
3
3
7
2
125 - 10y22 125 - 10y22 125 - 10y22 d `
15
375
13125
0

= 23.8 ft4

Ans.

Ans:
Ix = 23.8 ft 4
1018


© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

10–10.
Determine the moment of inertia for the shaded area about

the x axis.

y
y2

h2
—x
b
h

SOLUTION
d Ix =
Ix =

x

b

1 3
y dx
3

L

d Ix

=

b 3


b
y
1 h2 3>2 3>2
dx =
a b x dx
L0 3 b
L0 3

=

1 h2 3>2 2 5>2 b
a b a b x ]0
3 b
5

=

2
bh3
15

Ans.

Also,
dA = (b - x) dy = (b Ix =

L

y2 dA
h


=

L0

b 2
y ) dy
h2

y2 (b -

b 2
y ) dy
h2

b
b 5 h
= c y3 y d
3
5h2
0
=

2
bh3
15

Ans.

Ans:

Ix =
1019

2
bh3
15


© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

10–11.
y

Determine the moment of inertia for the shaded area about
the x axis.

8m
y ϭ 1 x3
8
x

Solution

4m
1

Differential Element. Here, x = 2y3. The area of the1 differential element parallel to
the x axis shown shaded in Fig. a is dA = xdy = 2y3 dy
Moment of Inertia. Perform the integration,



Ix =

LA

2

y dA =

L0

8m

L0



= 2



= 2a



1

2


y (2y3 dy)
8m

7

y3 dy

3 10 8 m
y3 b `
10
0

= 614.4 m4 = 614 m4

Ans.

Ans:
Ix = 614 m4
1020


© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

*10–12.
y

Determine the moment of inertia for the shaded area about
the y axis.


8m
y ϭ 1 x3
8
x

Solution

4m

Differential Element. The area of the differential element parallel to the y axis,
1
shown shaded in Fig. a, is dA = (8 - y)d x = a8 - x3 bdx
8

Moment of Inertia. Perform the integration,






Iy =

LA

x2dA =
=

L0


4m

L0

4m

x2 a8 a8x2 -

1 3
x b dx
8

1 5
x b dx
8

1 6 3
8
x b
= a x3 3
48

4m

0

= 85.33 m4 = 85.3 m4

Ans.


Ans:
Iy = 85.3 m4
1021


© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

10–13.
Determine the moment of inertia about the x axis.

y

x2 ϩ 4y2 ϭ 4

1m

x
2m

Solution
1
24 - x2. The moment of inertia of the differential
2
element parallel to the y axis shown shaded in Fig. a about x axis is

Differential Element. Here, y =




∼2
dIx = dIx′ + dAy



=



=



=



=

y 2
1
(dx)y3 + ydx a b
12
2
1 3
y dx
3

3
1 1

a 24 - x2 b dx
3 2

1
2(4 - x2)3 dx
24

Moment of Inertia. Perform the integration.


Ix =



=



=

L

dIx =

L0

2m

1
2(4 - x2)3 dx

24

1
x 2m
c x2(4 - x2)3 + 6x24 - x2 + 24 sin - 1 d `
96
2 0
p 4
m
8

Ans.

Ans:
Ix =
1022

p 4
m
8


© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

10–14.
y

Determine the moment of inertia about the y axis.


x2 ϩ 4y2 ϭ 4

1m

x
2m

Solution

1
24 - x2. The area of the differential element
2
1
parallel to the y axis shown shaded in Fig. a is dA = ydx = 24 - x2dx
2

Differential Element. Here, y =

Moment of Inertia. Perform the integration,


Iy =

LA

x2dA =



=




=



=

L0

2m

1
2 L0

1
x2 c 24 - x2dx d
2

2m

x2 24 - x2dx

2m
1 x
1
x
c - 2(4 - x2)3 + ax24 - x2 + 4 sin - 1 b d `
2 4

2
2
0

p 4
m
2



Ans.

Ans:
Iy =
1023

p 4
m
2


© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

10–15.
y

Determine the moment of inertia for the shaded area about
the x axis.
4 in.


y2 ϭ x
x
16 in.

Solution
Differential Element. The area of the differential element parallel with the x axis
shown shaded in Fig. a is dA = x dy = y2 dy.
Moment of Inertia. Perform the integration,


Ix =

LA

y2dA =

L0

4 in.

L0

4 in.

y2(y2dy)
y4dy




=



=



= 204.8 in4 = 205 in4

y5
5

`

4 in.
0

Ans.

Ans:
Ix = 205 in4
1024


© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

*10–16.
y


Determine the moment of inertia for the shaded area about
the y axis.
4 in.

y2 ϭ x
x
16 in.

Solution
Differential Element. The moment of inertia of the differential element parallel to
the x axis shown shaded in Fig. a about the y axis is
∼2
dIy = dIy + dAx



1
x 2
(dy)x3 + (xdy) a b
12
2



=


=


1 3
x dy
3



=

1 23
(y ) dy
3



=

1 6
y dy
3

Moment of Inertia. Perform the integration,


Iy =

L

dIy =

L0


4 in.

1 6
y dy
3

1 y7 4 in.
a b`
3 7 0



=

 

= 780.19 in4 = 780 in4

Ans.

Ans:
Iy = 780 in4
1025


© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

10–17.

y

Determine the moment of inertia for the shaded area about
the x axis.

h

y ϭ h x3
b3

x
b

Solution
Differential Element. The moment of inertia of the differential element parallel to
the y axis shown shaded in Fig. a about the x axis is


∼2
dIx = dIx′ + dAy



=



=




=



=

y 2
1
(dx)y3 + ydx a b
12
2
1 3
y dx
3

1 h 3 3
a x b dx
3 b3
h3 9
x dx
3b9

Moment of Inertia. Perform the integration,


Ix =

L


b

dIx =



=



=

h3 9
x dx
9
L0 3b
h3 x10 b
a b`
3b9 10 0
1 3
bh
30

Ans.



Ans:
Ix =
1026


1
bh3
30


© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

10–18.
y

Determine the moment of inertia for the shaded area about
the y axis.

h

y ϭ h x3
b3

x
b

Solution
Differential Element. The area of the differential element parallel to the y axis
h
shown shaded in Fig. a is dA = ydx = 3 x3dx
b
Moment of Inertia. Perform the integration,
Iy =


LA

x2dA =

L0

b

x2 a
b



=



=



=

h 3
x bdx
b3

h
x5dx

b3 L0
h x6 6
a b`
b3 6 0

b3h

6

Ans.

Ans:
Iy =
1027

b3h
6


© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

10–19.
y

Determine the moment of inertia for the shaded area about
the x axis.

y2 ϭ 1 Ϫ x
1m

x
1m
1m

Solution
1

Differential Element. Here y = (1 - x)2 . The moment of inertia of the
differential element parallel to the y axis shown shaded in Fig. a about the x axis is
3
1
1
2
2
2
dIx =
(dx)(2y)3 = y3dx = 3(1 - x)2 4 3 dx = (1 - x)2 dx.
12
3
3
3
Moment of Inertia. Perform the integration,
Ix =

L

dIx =




=



=

L0

1m

3
2
(1 - x)2 dx
3

1m
5
2 2
c - (1 - x)2 d `
3 5
0

4 4
m = 0.267 m4
15

Ans.

Ans:
Ix = 0.267 m4

1028


© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

*10–20.
y

Determine the moment of inertia for the shaded area about
the y axis.

y2 ϭ 1 Ϫ x
1m
x
1m
1m

Solution
Differential Element. Here x = 1 - y2. The moment of inertia of the differential
element parallel to the x axis shown shaded in Fig. a about the y axis is
∼2
dIy = dIy′ + dAx


=



=




=

1
x 2
(dy)x3 + xdy a b
12
2
1 3
x dy
3

1
(1 - y2)3dy
3

1
( -y6 + 3y4 - 3y2 + 1)dy
3
Moment of Inertia. Perform the integration,





=

Iy =


L

1m

dIy =



=



=

1
( - y6 + 3y4 - 3y2 + 1)dy
L-1 m 3
1m
3
1 y7
( - + y5 - y3 + y) `
3 7
5
-1 m

32 4
m = 0.305 m4
105


Ans.

Ans:
Iy = 0.305 m4
1029


© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

10–21.
y

Determine the moment of inertia for the shaded area about
the x axis.

y2 ϭ 2x
yϭx

2m

2m

x

Solution

1 2
y . The area of the differential element
2

1
parallel to the x axis shown shaded in Fig. a is dA = (x2 - x1)dy = ay - y2 bdy.
2
Moment of Inertia. Perform the integration,
Differential Element. Here x2 = y and x1 =

Ix =

LA

y2dA =

L0

L0



=



= a



2m

y2 ay -


2m

ay3 -

1 2
y bdy
2

1 4
y bdy
2

y5 2 m
y4
b2
4
10 0

= 0.8 m4

Ans.

Ans:
Ix = 0.8 m4
1030


© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


10–22.
Determine the moment of inertia for the shaded area about
the y axis.

y
y2 ϭ 2x
yϭx

2m

2m

x

Solution
1

Differential Element. Here, y2 = 22x2 and y1 = x. The area of the differential
element parallel to the y axis shown shaded in Fig. a is dA = (y2 - y1) dx
1
= 1 22x2 - x 2 dx.

Moment of Inertia. Perform the integration,
Iy =

LA

x2dA =

L0


2m

L0

2m



=



= a



=

1

x2 a 22x2 - xbdx

1 22x

5
2

- x3 2 dx


222 7
x4 2 2 m
x2 b
7
4 0

4 4
m = 0.571 m4
7

Ans.

Ans:
Iy = 0.571 m4
1031


© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

10–23.
y

Determine the moment of inertia for the shaded area about
the x axis.

b2
y2 ϭ —x
a
b x2

y ϭ—
a2

b

x

a

Solution

a 1
a
y2 and x1 = 2 y2. Thus, the area of the
1
b
b2
differential element parallel to the x axis shown shaded in Fig. a is dA = (x2 - x1) dy
Differential Element. Here x2 =

a
a 1
= a 1 y2 - 2 y2 b dy.
b
2
b

Moment of Inertia. Perform the integration,
Ix =


LA

b

y2dA =

a 1
a
y2 a 1 y2 - 2 y2 bdy
b
2
L0
b
L0



=



= a



=

b

a 5

a
a 1 y2 - 2 y4 bdy
b
2
b

2a

1
2

7b

7

y2 -

3ab3

35

a 5 2b
y b
5b2
0

Ans.

Ans:
Ix =

1032

3ab3
35


© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

*10–24.
Determine the moment of inertia for the shaded area about
the y axis.

y
b2
y2 ϭ —x
a
b x2
y ϭ—
a2

b

x

a

Solution

b 2

x . Thus, the area of the
a2
a
differential element parallel to the y axis shown shaded in Fig. a is dA = (y2 - y1)dx
b 1
b
= a 1 x2 - 2 x2 b dx
a
a2

Differential Element. Here, y2 =

b

1
2

1

x2 and y1 =

Moment of Inertia. Perform the integration,
Iy =

LA

a

x2dA =


b
b 1
x2 a 1 x2 - 2 x2 bdx
a
L0
a2
a

=

b 5
b
a 1 x2 - 2 x4 bdx
a
2
L0 a

= a

2b

7a

1
2

7

x2 -


b 5 2a
x b
5a2
0

=

2 3
1
a b - a3b
7
5

=

3a3b

35

Ans.

Ans:
Iy =
1033

3a3b
35


© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

10–25.
Determine the moment of inertia of the composite area
about the x axis.

y
3 in.

SOLUTION

3 in.

Composite Parts: The composite area can be subdivided into three segments as
shown in Fig. a. The perpendicular distance measured from the centroid of each
segment to the x axis is also indicated.

3 in.

6 in.

x

Moment of Inertia: The moment of inertia of each segment about the x axis can be
determined using the parallel-axis theorem. Thus,
-

Ix = Ix¿ + A(dy)2
= B


1
1
1
(3)(33) + (3)(3)(4)2 R + B (3)(33) + 3(3)(1.5)2 R
36
2
12

+ B

1
1
(6)(63) + (6)(6)(2)2 R
36
2

= 209 in4

Ans.

Ans:
Ix = 209 in4
1034