Uploaded by saba rabiee

attachment(۳)

PROBLEM 5.40
Determine by direct integration the centroid of the area shown. Express
your answer in terms of a and b.
SOLUTION
y1 = k1 x 2
but b = k1a 2
y2 = k 2 x 4
but b = k2 a 4
dA = ( y2 − y1 )dx =
b 2
x
a2
b
y2 = 4 x 4
a
y1 =
b
x4
2
x
−
dx
a2
a2
xEL = x
yEL =
=
1
( y1 + y2 )
2
b
x4
2
+
x
2a 2
a2
A = dA =
b
a2
a
0
b x3
x5
= 2
− 2
a 3 5a
x2 −
x4
dx
a2
a
0
2
= ba
15
xEL dA =
a
0
x
b
x4
2
−
x
dx
a2
a2
x5
dx
a2
=
b
a2
=
b x4
x6
−
a 2 4 6a 2
a
0
x3 −
a
0
1
= a 2b
12
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
596
PROBLEM 5.40 (Continued)
yEL dA =
=
a
0
b
x4 b
x4
2
2
+
−
x
x
dx
a2 a2
a2
2a 2
b2
2a 4
a
0
x4 −
x8
dx
a4
b 2 x5
x9
= 4
− 4
2 a 5 9a
a
=
0
2 2
ab
45
xA = xEL dA: x
2
1
ba = a 2b
15
12
5
x= a
8
yA = yEL dA: y
2
2 2
ba =
ab
15
45
1
y= b
3
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
597
PROBLEM 5.41
Determine by direct integration the centroid of the area shown. Express your
answer in terms of a and b.
SOLUTION
y =b
First note that symmetry implies
At
x = a,
y=b
y1 : b = ka 2
Then
y1 =
or
k=
b
a2
b 2
x
a2
y2 : b = 2b − ca 2
or
c=
b
a2
x2
a2
Then
y2 = b 2 −
Now
dA = ( y2 − y1 )dx2 = b 2 −
= 2b 1 −
and
Then
and
x2
b
− 2 x 2 dx
2
a
a
x2
dx
a2
xEL = x
A = dA
xEL dA =
a
0
a
0
x2
x3
2b 1 − 2 dx = 2b x − 2
a
3a
x2
x2
x4
x 2b 1 − 2 dx = 2b
− 2
2 4a
a
4
1
xA = xEL dA: x
ab = a 2b
3
2
a
=
0
4
ab
3
a
=
0
1 2
a b
2
3
x= a
8
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
598