PROBLEM 9.10
Determine the moment of inertia and the radius of gyration of the
shaded area shown with respect to the x axis.
SOLUTION Continued
Find:
and
xx
Ik
We have
33
21
333
3
1 1 12
( 2 ) sin
33 3 2
8( 2 ) sin
32
x
h
dI y y dx x a h x dx
aa
hxa x
a
a
π
π
= − = −+ −
= −+ −
Then
3
233
3
8( 2 ) sin
32
a
xx
a
h
I dI x a x dx
a
a
π
= = −+ −
∫∫
Now
32 2
sin sin (1 cos ) sin sin cos
θ θ θ θ θθ
=−=−
Then
3232
3
2
343
3
34
3
3
8( 2 ) sin sin cos
3 2 22
2 22
( 2 ) cos cos
3 23 2
22 2
( 2)
33
22
1
33
a
xa
a
a
h
I x a x x x dx
a aa
a
h aa
xa x x
aa
a
h aa aa
a
ah
π ππ
ππ
ππ
ππ
π
= −+ − −
= − −+ + −
= − + + −+
= −
∫
3
0.52520
x
I ah=
3
or 0.525
x
I ah=
and
3
20.52520
0.36338
x
xIah
kA ah
= =
or
1.202
x
kh=
PROBLEM 7.11
Determine the moment of inertia and the radius of gyration of the shaded
area shown with respect to the x axis.
SOLUTION
At
11 2
,:xay y b= = =
2
12
2
22
: or
: 2 or
b
y b ka k ab
y b b ca c a
= =
=−=
Then
2
2
12
22
2
bx
y xyb
aa
= = −
Now
22 22
21 22 2
2
() 2 ()
xb b
dA y y dx b x dx a x dx
aa a
=− = −− = −
Then
22 2 3
22
00
2 2 14
() 33
a
a
bb
A dA a x dx a x x ab
aa
== −= − =
∫∫
Now
33
2
33 2
21 22
36 42 24 6 6
6
36 42 24 6
6
11 12
33 3
1(8 12 6 )
3
2(4 6 3 )
3
x
xb
dI y y dx b x dx
aa
ba ax ax x x dx
a
ba ax ax x dx
a
=− = −−
= − + −−
= −+−
Then
36 42 24 6
6
0
36 43 65 7
60
2(4 6 3 )
3
2 31
42
3 57
a
xx
a
b
I dI a a x a x x dx
a
bax ax ax x
a
== −+−
= −+ −
∫∫
3
172
105 ab=
3
or 1.638
x
I ab=
and
3
172
22
105
4
3
43
35
x
x
ab
I
kb
A ab
= = =
or 1.108
x
kb=
PROBLEM 7.12
Determine the moment of inertia and the radius of gyration of the
shaded area shown with respect to the x axis.
SOLUTION
2 1/2
11 2 2
y kx y kx= =
For
12
0 andx yyb
= = =
2 1/2
12
12
2 1/2
bka bka
bb
kk
aa
= =
= =
Thus,
2 1/2
12
2 1/2
bb
y xy x
aa
= =
21
1/2 2
1/2 2
0
3/2 3
1/2 2
()
2
33
1
3
a
dA y y dx
bb
A x x dx
aa
ba ba
Aaa
A ab
= −
= −
= −
=
∫
33
21
33
3/2 6
3/2 6
11
33
11
33
x
dI y dx y dx
bb
x dx x dx
aa
= −
= −
()
33
3/2 6
3/2 6
00
3 5/2 3 7 3
3/2 6
5
2
33
21
7 15 21
33
aa
xx
bb
I dI x dx x dx
aa
b a ba ab
aa
= = −
= −=−
∫∫ ∫
3
3
35
x
I ab=
( )
3
3
35
2x
xab
b
ab
I
kA
= =
9
35
x
kb
=
consent of McGraw–Hill Education.
PROBLEM 7.13
Determine the moment of inertia and the radius of gyration of the shaded area
shown with respect to the y axis.
SOLUTION
22
22
2
2
22
1
1
2
2
y
xy
ab
x
yb a
dA ydx
dI x dA x ydx
+=
= −
=
= =
2
22
2
00
2 21
aa
yy
x
I dI x ydx b x dx
a
= = = −
∫∫ ∫
Set:
sin cosxa dxa d
θ θθ
= =
/2 22 2
0
/2 /2
3 22 3 2
00
/2
/2
33
00
2 sin 1 sin cos
1
2 sin cos 2 sin 2
4
11 11
(1 cos 4 ) sin 4
22 44
y
I ba a d
ab d ab d
ab d ab
π
ππ
π
π
θ θ θθ
θ θθ θθ
θθ θ θ
= −
= =
= −=−
∫
∫∫
∫
33
10
42 8
ab ab
ππ
= −=
3
1
8
y
I ab
π
=
From solution of Problem 7.9:
1
2
A ab
π
=
Thus:
3
1
22 2
8
1
2
1
4
y
yy y
Iab
I kA k a
A ab
π
π
= = = =
1
2
y
ka=
PROBLEM 7.14
Determine the moment of inertia and the radius of gyration of the
shaded area shown with respect to the y axis.
SOLUTION
1:y
At
2 , 0:x ay= =
0 sin (2 )c ka=
2 or 2
ak k a
π
π
= =
At
,:xa yh= =
sin ( )
2
hc a
a
π
=
or
ch=
2:
y
At
, 2:xa y h= =
2h ma b= −
At
2 , 0:x ay= =
0 (2 )ma b= +
Solving yields
2,4
h
m bh
a
=−=
Then
12
2
sin 4
22( 2)
h
yh x y xh
aa
hxa
a
π
= =−+
= −+
Now
21 2
( ) ( 2 ) sin 2
h
dA y y dx x a h x dx
aa
π
= − = −+ −
Then
2
2( 2 ) sin 2
a
a
A dA h x a x dx
aa
π
= = −+ −
∫∫
2
2
12
( 2 ) cos 2
a
a
a
h xa x
aa
π
π
= − −+ +
2
21 2
( 2) 1
0.36338
a
h a a ah
a
ah
ππ
= − + −+ = −
=
SOLUTION Continued
Find:
and
yy
Ik
We have
22
3 22
2( 2 ) sin 2
2( 2 ) sin 2
y
h
dI x dA x x a h x dx
aa
h x ax x x dx
aa
π
π
= = −+ −
= −+ −
Then
23 22
2( 2 ) sin 2
a
yy
a
I dI h x ax x x dx
aa
π
= = −+ −
∫∫
Now using integration by parts with
2sin 2
u x dv xdx
a
π
= =
2
2 cos 2
a
du xdx v x
a
π
π
= = −
Then
22
22
sin cos cos (2 )
2 22
aa
x xdx x x x xdx
a aa
π ππ
ππ
= − −−
∫∫
Now let
cos 2
u x dv xdx
a
π
= =
2sin 2
a
du dx v x
a
π
π
= =
Then
22
2 42 2
sin cos sin sin
2 2 22
a aa a
x xdx x x x x x dx
a a aa
π π ππ
π ππ π
=−+ −
∫∫
43
2
23
223
21 2
43
2 8 16
cos sin cos
222
y
a
a
I h x ax
a
aa a
x xxx x
aaa
πππ
πππ
= −+
−− + +
3
43 2
3
2
43
2
3
2 1 2 2 16
(2 ) (2 ) (2 )
43
21 2 8
() () ()
43
0.61345
aa
h a aa a
a
a
h a aa a
a
ah
ππ
π
=−+ − +
−− + −
=
or
3
0.613
y
I ah=
and
3
20.61345
0.36338
y
y
Iah
kA ah
= =
or
1.299
y
ka=
consent of McGraw–Hill Education.
PROBLEM 7.15
Determine the moment of inertia and the radius of gyration of the shaded
area shown with respect to the y axis.
SOLUTION
At
12
,:xa y y b
= = =
2
12
2
22
: or
: 2 or
b
y b ka k ab
y b b ca c a
= =
=−=
Then
2
12
2
22
2
b
yx
a
x
yb a
=
= −
Now
21
22
22
22
2
()
2
2()
dA y y dx
xb
b x dx
aa
ba x dx
a
= −
= −−
= −
Then
22
2
0
23
20
2()
21
3
4
3
a
a
b
A dA a x dx
a
bax x
a
ab
= = −
= −
=
∫∫
Now
2 2 22
2
2()
y
b
dI x dA x a x dx
a
= = −
consent of McGraw–Hill Education.
SOLUTION Continued
Then
22 2
2
0
23 5
20
2()
21 1
35
a
yy
a
b
I dI x a x dx
a
bax x
a
= = −
= −
∫∫
or
3
4
15
y
I ab=
and
3
4
22
15
4
3
1
5
y
y
Iab
ka
A ab
= = =
or
5
ya
k=
consent of McGraw–Hill Education.
PROBLEM 7.16
Determine the moment of inertia and the radius of gyration of the
shaded area shown with respect to the y axis.
SOLUTION
See figure of solution on Problem 7.12.
22
21
2 1/2 2 5/2 4
1/2 2 1/2 2
0 00
1()
3y
a aa
y
A ab dI x dA x y y dx
bb b b
I x x x dx x dx x dx
aa a a
= = = −
= −= −
∫ ∫∫
( )
7/2 5 3
1/2 2
7
2
21
5 75
yb b ba
I ab
aa
= ⋅ −⋅=−
3
3
35
y
I ab=
( )
3
3
35
2
3
y
yab
ab
I
kA
= =
9
35
y
ka=
consent of McGraw–Hill Education.
SOLUTION Continued
Find:
and
xx
Ik
We have
33
21
333
3
1 1 12
( 2 ) sin
33 3 2
8( 2 ) sin
32
x
h
dI y y dx x a h x dx
aa
hxa x
a
a
π
π
= − = −+ −
= −+ −
Then
3
233
3
8( 2 ) sin
32
a
xx
a
h
I dI x a x dx
a
a
π
= = −+ −
∫∫
Now
32 2
sin sin (1 cos ) sin sin cos
θ θ θ θ θθ
=−=−
Then
3232
3
2
343
3
34
3
3
8( 2 ) sin sin cos
3 2 22
2 22
( 2 ) cos cos
3 23 2
22 2
( 2)
33
22
1
33
a
xa
a
a
h
I x a x x x dx
a aa
a
h aa
xa x x
aa
a
h aa aa
a
ah
π ππ
ππ
ππ
ππ
π
= −+ − −
= − −+ + −
= − + + −+
= −
∫
3
0.52520
x
I ah=
3
or 0.525
x
I ah=
and
3
20.52520
0.36338
x
xIah
kA ah
= =
or
1.202
x
kh=
PROBLEM 7.11
Determine the moment of inertia and the radius of gyration of the shaded
area shown with respect to the x axis.
SOLUTION
At
11 2
,:xay y b= = =
2
12
2
22
: or
: 2 or
b
y b ka k ab
y b b ca c a
= =
=−=
Then
2
2
12
22
2
bx
y xyb
aa
= = −
Now
22 22
21 22 2
2
() 2 ()
xb b
dA y y dx b x dx a x dx
aa a
=− = −− = −
Then
22 2 3
22
00
2 2 14
() 33
a
a
bb
A dA a x dx a x x ab
aa
== −= − =
∫∫
Now
33
2
33 2
21 22
36 42 24 6 6
6
36 42 24 6
6
11 12
33 3
1(8 12 6 )
3
2(4 6 3 )
3
x
xb
dI y y dx b x dx
aa
ba ax ax x x dx
a
ba ax ax x dx
a
=− = −−
= − + −−
= −+−
Then
36 42 24 6
6
0
36 43 65 7
60
2(4 6 3 )
3
2 31
42
3 57
a
xx
a
b
I dI a a x a x x dx
a
bax ax ax x
a
== −+−
= −+ −
∫∫
3
172
105 ab=
3
or 1.638
x
I ab=
and
3
172
22
105
4
3
43
35
x
x
ab
I
kb
A ab
= = =
or 1.108
x
kb=
PROBLEM 7.12
Determine the moment of inertia and the radius of gyration of the
shaded area shown with respect to the x axis.
SOLUTION
2 1/2
11 2 2
y kx y kx= =
For
12
0 andx yyb
= = =
2 1/2
12
12
2 1/2
bka bka
bb
kk
aa
= =
= =
Thus,
2 1/2
12
2 1/2
bb
y xy x
aa
= =
21
1/2 2
1/2 2
0
3/2 3
1/2 2
()
2
33
1
3
a
dA y y dx
bb
A x x dx
aa
ba ba
Aaa
A ab
= −
= −
= −
=
∫
33
21
33
3/2 6
3/2 6
11
33
11
33
x
dI y dx y dx
bb
x dx x dx
aa
= −
= −
()
33
3/2 6
3/2 6
00
3 5/2 3 7 3
3/2 6
5
2
33
21
7 15 21
33
aa
xx
bb
I dI x dx x dx
aa
b a ba ab
aa
= = −
= −=−
∫∫ ∫
3
3
35
x
I ab=
( )
3
3
35
2x
xab
b
ab
I
kA
= =
9
35
x
kb
=
consent of McGraw–Hill Education.
PROBLEM 7.13
Determine the moment of inertia and the radius of gyration of the shaded area
shown with respect to the y axis.
SOLUTION
22
22
2
2
22
1
1
2
2
y
xy
ab
x
yb a
dA ydx
dI x dA x ydx
+=
= −
=
= =
2
22
2
00
2 21
aa
yy
x
I dI x ydx b x dx
a
= = = −
∫∫ ∫
Set:
sin cosxa dxa d
θ θθ
= =
/2 22 2
0
/2 /2
3 22 3 2
00
/2
/2
33
00
2 sin 1 sin cos
1
2 sin cos 2 sin 2
4
11 11
(1 cos 4 ) sin 4
22 44
y
I ba a d
ab d ab d
ab d ab
π
ππ
π
π
θ θ θθ
θ θθ θθ
θθ θ θ
= −
= =
= −=−
∫
∫∫
∫
33
10
42 8
ab ab
ππ
= −=
3
1
8
y
I ab
π
=
From solution of Problem 7.9:
1
2
A ab
π
=
Thus:
3
1
22 2
8
1
2
1
4
y
yy y
Iab
I kA k a
A ab
π
π
= = = =
1
2
y
ka=
PROBLEM 7.14
Determine the moment of inertia and the radius of gyration of the
shaded area shown with respect to the y axis.
SOLUTION
1:y
At
2 , 0:x ay= =
0 sin (2 )c ka=
2 or 2
ak k a
π
π
= =
At
,:xa yh= =
sin ( )
2
hc a
a
π
=
or
ch=
2:
y
At
, 2:xa y h= =
2h ma b= −
At
2 , 0:x ay= =
0 (2 )ma b= +
Solving yields
2,4
h
m bh
a
=−=
Then
12
2
sin 4
22( 2)
h
yh x y xh
aa
hxa
a
π
= =−+
= −+
Now
21 2
( ) ( 2 ) sin 2
h
dA y y dx x a h x dx
aa
π
= − = −+ −
Then
2
2( 2 ) sin 2
a
a
A dA h x a x dx
aa
π
= = −+ −
∫∫
2
2
12
( 2 ) cos 2
a
a
a
h xa x
aa
π
π
= − −+ +
2
21 2
( 2) 1
0.36338
a
h a a ah
a
ah
ππ
= − + −+ = −
=
SOLUTION Continued
Find:
and
yy
Ik
We have
22
3 22
2( 2 ) sin 2
2( 2 ) sin 2
y
h
dI x dA x x a h x dx
aa
h x ax x x dx
aa
π
π
= = −+ −
= −+ −
Then
23 22
2( 2 ) sin 2
a
yy
a
I dI h x ax x x dx
aa
π
= = −+ −
∫∫
Now using integration by parts with
2sin 2
u x dv xdx
a
π
= =
2
2 cos 2
a
du xdx v x
a
π
π
= = −
Then
22
22
sin cos cos (2 )
2 22
aa
x xdx x x x xdx
a aa
π ππ
ππ
= − −−
∫∫
Now let
cos 2
u x dv xdx
a
π
= =
2sin 2
a
du dx v x
a
π
π
= =
Then
22
2 42 2
sin cos sin sin
2 2 22
a aa a
x xdx x x x x x dx
a a aa
π π ππ
π ππ π
=−+ −
∫∫
43
2
23
223
21 2
43
2 8 16
cos sin cos
222
y
a
a
I h x ax
a
aa a
x xxx x
aaa
πππ
πππ
= −+
−− + +
3
43 2
3
2
43
2
3
2 1 2 2 16
(2 ) (2 ) (2 )
43
21 2 8
() () ()
43
0.61345
aa
h a aa a
a
a
h a aa a
a
ah
ππ
π
=−+ − +
−− + −
=
or
3
0.613
y
I ah=
and
3
20.61345
0.36338
y
y
Iah
kA ah
= =
or
1.299
y
ka=
consent of McGraw–Hill Education.
PROBLEM 7.15
Determine the moment of inertia and the radius of gyration of the shaded
area shown with respect to the y axis.
SOLUTION
At
12
,:xa y y b
= = =
2
12
2
22
: or
: 2 or
b
y b ka k ab
y b b ca c a
= =
=−=
Then
2
12
2
22
2
b
yx
a
x
yb a
=
= −
Now
21
22
22
22
2
()
2
2()
dA y y dx
xb
b x dx
aa
ba x dx
a
= −
= −−
= −
Then
22
2
0
23
20
2()
21
3
4
3
a
a
b
A dA a x dx
a
bax x
a
ab
= = −
= −
=
∫∫
Now
2 2 22
2
2()
y
b
dI x dA x a x dx
a
= = −
consent of McGraw–Hill Education.
SOLUTION Continued
Then
22 2
2
0
23 5
20
2()
21 1
35
a
yy
a
b
I dI x a x dx
a
bax x
a
= = −
= −
∫∫
or
3
4
15
y
I ab=
and
3
4
22
15
4
3
1
5
y
y
Iab
ka
A ab
= = =
or
5
ya
k=
consent of McGraw–Hill Education.
PROBLEM 7.16
Determine the moment of inertia and the radius of gyration of the
shaded area shown with respect to the y axis.
SOLUTION
See figure of solution on Problem 7.12.
22
21
2 1/2 2 5/2 4
1/2 2 1/2 2
0 00
1()
3y
a aa
y
A ab dI x dA x y y dx
bb b b
I x x x dx x dx x dx
aa a a
= = = −
= −= −
∫ ∫∫
( )
7/2 5 3
1/2 2
7
2
21
5 75
yb b ba
I ab
aa
= ⋅ −⋅=−
3
3
35
y
I ab=
( )
3
3
35
2
3
y
yab
ab
I
kA
= =
9
35
y
ka=
consent of McGraw–Hill Education.