Chapter 9 Solution Manual: Probability, Statistics & Random Processes – Problem Answers & Explanations

CHAPTER9
j

, PROBLEMj9.1

Determinejbyjdirectjintegrationjthejmomentjofjinertiajofjthejshadedjareajwit
hjrespectjtojthejyjaxis.




SOLUTION




Atj xj=j0,j j yj=jbj:j j bj=jkj(0j−ja)
2


b bj
kj =j then yj=j (jxj−ja)2j;jdAj=j ydx
2
a a2

dIjyj =j x2dAj=j x2jyjdx
ajj b 2j
x 2 xj−ja
Ijyj =j  0j ja2 dx

aj j b 
x 4 −j2ax 3+jaj x2

2
=j j
0j ja2 dx

bj j1 5 1 4 1 2j 3ja
= x j − ax j + a x 
a2j j5 2 3 0
1j
Ijyj =j a3bj ◀
30




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, PROBLEMj9.2

Determinej byj directj integrationj thej momentjofj inertiajofj thej shadedj areajw
ithjrespectjtojthejyjaxis.




SOLUTION

yj=jkx1/3

Forj xj=ja:



bj 1/3
Thus: yj= x
a1/3

dIjyj =j x2dAj=j x2jydx

b 1/3 bj
dIy =jx2 1/3
x dxj= 1/3 x7/3dx
a a
b a b j 3  3j

Iy =j j dIj jy= 1/3
a  0
x7/3dxj = 1/3  a10/3 
a  10 
Ijyj =j
10
a3bj◀




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, PROBLEMj9.3

Determinej byj directj integrationj thej momentjofj inertiajofj thej shadedj areajw
ithjrespectjtojthejyjaxis.




SOLUTION

hj
Byjobservation yj=j xj
b
jNow dIjyj =jx2 dAj=j x2[(hj−j y)dx]
 xj
=jhx 2 1j−  dx
 bj 
b  xj
Then 
Ijyj =j j dIj jy=  0
hx 2 1j− jdx
 bj
b
j1 3 x4j  1 3
=jhjj j x j −  or Ijyj = b hj◀
j3 4bj0 12




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