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PROBLEM 7.2
Determine by direct integration the moment of inertia of the shaded area with
respect to the y axis.
SOLUTION
1/3
y kx=
For
:xa=
1/3
b ka=
1/3
/
k ba
=
Thus:
1/3
1/3
b
yx
a
=
22
y
dI x dA x ydx= =
2 1/3 7/3
1/3 1/3
7/3 10/3
1/3 1/3
0
3
10
y
a
yy
bb
dI x x dx x dx
aa
bb
I dI x dx a
aa
= =
= = =
∫∫
3
3
10
y
I ab=
consent of McGraw-Hill Education.
PROBLEM 7.3
Determine by direct integration the moment of inertia of the shaded area
with respect to the y axis.
SOLUTION
2
2
4xx
yh
aa
= −
2
22
2
34
2
0
4
4
y
a
y
dA ydx
xx
dI x dA hx dx
aa
xx
I h dx
aa
=
= = −
= −
∫
4 5 33
20
44
4 45
5
a
y
x x aa
Ih h
aa
= −= −
3
1
5
y
I ha=
consent of McGraw-Hill Education.
PROBLEM 7.4
Determine by direct integration the moment of inertia of the shaded area
with respect to the y axis.
SOLUTION
At
1 12
,xayyb= = =
1
:y
22
or b
b ka k a
= =
2:y
or b
b ma m a
= =
Then
2
12
b
yx
a
=
2b
yx
a
=
Now
22 2 2
21 2
[( )
y
bb
dI x dA x y y dx x x x dx
aa
== −= −
Then
34
2
0
45
20
11
11
45
a
yy
a
I dI b x x dx
aa
bx x
aa
= = −
= −
∫∫
or
3
1
20
y
I ab
=
consent of McGraw-Hill Education.
PROBLEM 7.5
Determine by direct integration the moment of inertia of the shaded area with
respect to the x axis.
SOLUTION
3
1 21
1
() 3
x
x
y h h h dI y dx
a
=+− =
( )
3
1 21
0
4
1 21 21
0
22
212121
44
21
21 21
1()
3
1()
12
( )( )( )
12( ) 12
a
xy
a
x
I dI h h h dx
a
xa
h hh
a hh
hhhhhh
aa
hh
hh hh
= = +−
= +−
−
++−
= −=⋅
−−
∫∫
22
1 21 2
( )( )
12
x
a
I h hhh=++
consent of McGraw-Hill Education.
PROBLEM 7.6
Determine by direct integration the moment of inertia of the shaded area
with respect to the x axis.
SOLUTION
1/3
1/3
b
yx
a
=
33
3 1/3
1/3
3 32
0
11 1
33 3
11
3 32
x
a
xx
bb
dI y dx x dx xdx
a
a
b ba
I dI xdx
aa
= = =
= = =
∫∫
3
1
6
x
I ab=
consent of McGraw-Hill Education.
PROBLEM 7.7
Determine by direct integration the moment of inertia of the shaded
area with respect to the x axis.
SOLUTION
2
2
3
2
32
3 3 4 56
3 4 56
0
34 5 6 7
3456
0
33
4
11
4
33
64 33
3
64 1 3 1 1
34 5 2 7
64 1 3 1 1 64 3
3 4 5 2 7 3 420
x
a
xx
a
xx
yh
aa
xx
dI y dx h dx
aa
h x x xx
I dI dx
a a aa
hx x x x
aaaa
hh
aa
= −
= = −
= = −+−
= −+−
= −+− =
∫∫
3
16
105
x
I ah=
PROBLEM 7.8
Determine by direct integration the moment of inertia of the shaded area
with respect to the x axis.
SOLUTION
At
1 12
,xayyb= = =
1:y
22
or b
b ka k a
= =
2:
y
or b
b ma m a
= =
Then
2
12
b
yx
a
=
2b
yx
a
=
Now
( )
33
21
33
36
36
1
3
1
3
x
dI y y dx
bb
x x dx
aa
= −
= −
Then
336
36
0
347
36
0
11
3
11
347
a
xx
a
b
I dI x x dx
aa
bxx
aa
= = −
= −
∫∫
or
3
1
28
x
I ab=
consent of McGraw-Hill Education.
PROBLEM 7.9
Determine the moment of inertia and the radius of gyration of the shaded
area shown with respect to the x axis.
SOLUTION
22
22
2
2
22
1
1
x
xy
ab
y
xa b
dA xdy
dI y dA y xdy
+=
= −
=
= =
2
22
2
1
bb
xx
bb
y
I dI xy dy a y dy
b
−−
= = = −
∫∫ ∫
Set:
sin cosyb dyb d
θ θθ
= =
/2 22 2
/2
/2 /2
3 22 3 2
/2 /2
/2
/2
33
/2 /2
sin 1 sin cos
1
sin cos sin 2
4
11 11
(1 cos 4 ) sin 4
42 84
x
Ia b b d
ab d ab d
ab d ab
π
π
ππ
ππ
π
π
ππ
θ θ θθ
θ θθ θθ
θθ θ θ
−
−−
−−
= −
= =
= −=−
∫
∫∫
∫
32
1
82 28
ab ab
π ππ
= −− =
3
1
8
x
I ab
π
=
2/2 2
2/2
/2 /2
2
/2 /2
/2
/2
1 1 sin cos
1
cos (1 cos2 )
2
11
sin 2
2 22 2 2
bb
bb
y
A dA xdy a dy a b d
b
ab d ab d
ab
ab ab
π
π
ππ
ππ
π
π
θ θθ
θθ θ θ
ππ
θθ π
+
−− −
−−
−
== = −= −
= = +
= + = −− =
∫∫ ∫ ∫
∫∫
Problem 7.9 (Continued)
3
1
22 2
8
1
2
1
4
x
xx x
ab
I
I kA k b
A ab
π
π
= = = =
1
2
x
kb=
consent of McGraw-Hill Education.
PROBLEM 7.2
Determine by direct integration the moment of inertia of the shaded area with
respect to the y axis.
SOLUTION
1/3
y kx=
For
:xa=
1/3
b ka=
1/3
/
k ba
=
Thus:
1/3
1/3
b
yx
a
=
22
y
dI x dA x ydx= =
2 1/3 7/3
1/3 1/3
7/3 10/3
1/3 1/3
0
3
10
y
a
yy
bb
dI x x dx x dx
aa
bb
I dI x dx a
aa
= =
= = =
∫∫
3
3
10
y
I ab=
consent of McGraw-Hill Education.
PROBLEM 7.3
Determine by direct integration the moment of inertia of the shaded area
with respect to the y axis.
SOLUTION
2
2
4xx
yh
aa
= −
2
22
2
34
2
0
4
4
y
a
y
dA ydx
xx
dI x dA hx dx
aa
xx
I h dx
aa
=
= = −
= −
∫
4 5 33
20
44
4 45
5
a
y
x x aa
Ih h
aa
= −= −
3
1
5
y
I ha=
consent of McGraw-Hill Education.
PROBLEM 7.4
Determine by direct integration the moment of inertia of the shaded area
with respect to the y axis.
SOLUTION
At
1 12
,xayyb= = =
1
:y
22
or b
b ka k a
= =
2:y
or b
b ma m a
= =
Then
2
12
b
yx
a
=
2b
yx
a
=
Now
22 2 2
21 2
[( )
y
bb
dI x dA x y y dx x x x dx
aa
== −= −
Then
34
2
0
45
20
11
11
45
a
yy
a
I dI b x x dx
aa
bx x
aa
= = −
= −
∫∫
or
3
1
20
y
I ab
=
consent of McGraw-Hill Education.
PROBLEM 7.5
Determine by direct integration the moment of inertia of the shaded area with
respect to the x axis.
SOLUTION
3
1 21
1
() 3
x
x
y h h h dI y dx
a
=+− =
( )
3
1 21
0
4
1 21 21
0
22
212121
44
21
21 21
1()
3
1()
12
( )( )( )
12( ) 12
a
xy
a
x
I dI h h h dx
a
xa
h hh
a hh
hhhhhh
aa
hh
hh hh
= = +−
= +−
−
++−
= −=⋅
−−
∫∫
22
1 21 2
( )( )
12
x
a
I h hhh=++
consent of McGraw-Hill Education.
PROBLEM 7.6
Determine by direct integration the moment of inertia of the shaded area
with respect to the x axis.
SOLUTION
1/3
1/3
b
yx
a
=
33
3 1/3
1/3
3 32
0
11 1
33 3
11
3 32
x
a
xx
bb
dI y dx x dx xdx
a
a
b ba
I dI xdx
aa
= = =
= = =
∫∫
3
1
6
x
I ab=
consent of McGraw-Hill Education.
PROBLEM 7.7
Determine by direct integration the moment of inertia of the shaded
area with respect to the x axis.
SOLUTION
2
2
3
2
32
3 3 4 56
3 4 56
0
34 5 6 7
3456
0
33
4
11
4
33
64 33
3
64 1 3 1 1
34 5 2 7
64 1 3 1 1 64 3
3 4 5 2 7 3 420
x
a
xx
a
xx
yh
aa
xx
dI y dx h dx
aa
h x x xx
I dI dx
a a aa
hx x x x
aaaa
hh
aa
= −
= = −
= = −+−
= −+−
= −+− =
∫∫
3
16
105
x
I ah=
PROBLEM 7.8
Determine by direct integration the moment of inertia of the shaded area
with respect to the x axis.
SOLUTION
At
1 12
,xayyb= = =
1:y
22
or b
b ka k a
= =
2:
y
or b
b ma m a
= =
Then
2
12
b
yx
a
=
2b
yx
a
=
Now
( )
33
21
33
36
36
1
3
1
3
x
dI y y dx
bb
x x dx
aa
= −
= −
Then
336
36
0
347
36
0
11
3
11
347
a
xx
a
b
I dI x x dx
aa
bxx
aa
= = −
= −
∫∫
or
3
1
28
x
I ab=
consent of McGraw-Hill Education.
PROBLEM 7.9
Determine the moment of inertia and the radius of gyration of the shaded
area shown with respect to the x axis.
SOLUTION
22
22
2
2
22
1
1
x
xy
ab
y
xa b
dA xdy
dI y dA y xdy
+=
= −
=
= =
2
22
2
1
bb
xx
bb
y
I dI xy dy a y dy
b
−−
= = = −
∫∫ ∫
Set:
sin cosyb dyb d
θ θθ
= =
/2 22 2
/2
/2 /2
3 22 3 2
/2 /2
/2
/2
33
/2 /2
sin 1 sin cos
1
sin cos sin 2
4
11 11
(1 cos 4 ) sin 4
42 84
x
Ia b b d
ab d ab d
ab d ab
π
π
ππ
ππ
π
π
ππ
θ θ θθ
θ θθ θθ
θθ θ θ
−
−−
−−
= −
= =
= −=−
∫
∫∫
∫
32
1
82 28
ab ab
π ππ
= −− =
3
1
8
x
I ab
π
=
2/2 2
2/2
/2 /2
2
/2 /2
/2
/2
1 1 sin cos
1
cos (1 cos2 )
2
11
sin 2
2 22 2 2
bb
bb
y
A dA xdy a dy a b d
b
ab d ab d
ab
ab ab
π
π
ππ
ππ
π
π
θ θθ
θθ θ θ
ππ
θθ π
+
−− −
−−
−
== = −= −
= = +
= + = −− =
∫∫ ∫ ∫
∫∫
Problem 7.9 (Continued)
3
1
22 2
8
1
2
1
4
x
xx x
ab
I
I kA k b
A ab
π
π
= = = =
1
2
x
kb=
consent of McGraw-Hill Education.
