PROBLEM 5.31
Determine by direct integration the centroid of the area shown.
SOLUTION
For the element (EL) shown,
22
b
y ax
a
= −
and
()
()
22
22
()
1()
2
2
EL
EL
dA b y dx
ba a x dx
a
xx
y yb
b
a ax
a
= −
=−−
=
= +
= +−
Then
()
22
0
a
b
A dA a a x dx
a
== −−
∫∫
To integrate, let
22
sin : cos , cosxa a x a dxa d
θ θ θθ
= −= =
Then
/2
0
/2
22
0
( cos )( cos )
2
sin sin
24
14
b
A aa a d
a
baa
a
ab
π
π
θ θθ
θθ
θ
π
= −
= −+
= −
∫
and
()
22
0
/2
2 2 2 3/2
0
3
1()
23
1
6
a
EL
b
x dA x a a x dx
a
ba
x ax
a
ab
π
= −−
= +−
=
∫∫
SOLUTION Continued
()()
22 22
0
2 23
2
22
00
2
2
( ) 3
22
1
6
a
EL
a
a
bb
y dA a a x a a x dx
aa
b bx
x dx
aa
ab
= +− −−
= =
=
∫∫
∫
2
1
:1
46
EL
xA x dA x ab a b
π
= −=
∫
2
or 3(4 )
a
x
π
=−
2
1
:1
46
EL
yA y dA y ab ab
π
= −=
∫
2
or 3(4 )
b
y
π
=−
PROBLEM 5.32
Determine by direct integration the centroid of the area shown.
SOLUTION
2
21
,
x
y y kx
k
= =
But
2,a ka=
thus,
2
1
ka
=
2
21
,x
y ax y a
= =
2
21
2
0
3
3/2 2
0
()
21
3 33
EL
a
a
xx
x
dA y y dx ax dx
a
x
A dA ax dx
a
x
ax a
a
=
=−=−
= = −
= −=
∫∫
2 34
3/2 5/2 3
00 0
23
5 4 20
a
aa
EL
x xx
x dA x ax dx ax dx ax a
aaa
= − = − = −=
∫∫ ∫
23
13 9
:3 20 20
EL
a
xA x dA x a a x
= = =
∫
By symmetry,
9
20
a
yx= =
consent of McGraw–Hill Education.
PROBLEM 5.33
Determine by direct integration the centroid of the area shown. Express
your answer in terms of a and b.
SOLUTION
For y2 at
,xa=
22
, , or a
y b a kb k b
= = =
Then
1/2
2
b
yx
a
=
Now
EL
xx=
and for
0,
2
a
x≤≤
1/2
2
1/2
2
22
EL
ybx
ya
x
dA y dx b dx
a
= =
= =
For
,
2
axa
≤≤
1/2
12
11
()
2 22
EL bx x
y yy aa
= + = −+
1/2
21
1
() 2
xx
dA y y dx b dx
a
a
= − = −+
Then
1/2 1/2
/2
0 /2
1
2
aa
a
x xx
A dA b dx b dx
a
aa
= = + −+
∫∫ ∫
/2 3/2 2
3/2
0/2
3/2 3/2
3/2
2
2
2 21
3 3 22
2()
32 2
11
( ) ()
2 22 2
13
24
a
a
a
b xx
xb x
a
aa
ba a
a
a
aa
ba a
a
ab
= + −+
= +−
+− − + −
=
consent of McGraw–Hill Education.
SOLUTION Continued
and
1/2 1/2
/2
0 /2
1
2
aa
EL a
x xx
x dA x b dx x b dx
a
aa
= + −+
∫∫ ∫
/2 5/2 3 4
5/2
0/2
5/2 5/2
5/2
32
32
2
22
5 5 34
2()
52 2
11
() ()
3 24 2
71
240
a
a
a
b x xx
xb a
aa
ba a
a
a
aa
ba a
a
ab
= + −+
= +−
+− − + −
=
1/2 1/2
/2
0
1/2 1/2
/2
/2 3
2 22
2
0/2
22
2
2
2
11
22 2
1 11
22 2 2 3 2
1
()
4 2 2 6 22
a
EL
a
a
a
a
a
bx x
y dA b dx
aa
bx x x x
b dx
aa
aa
b bx x
x
a a aa
b a a ba
a
aa
=
+ −+ −+
= + −−
= +− − −
∫∫
∫
3
2
11
48
ab
=
Hence,
2
13 71
:24 240
EL
xA x dA x ab a b
= =
∫
17
130
xa=
2
13 11
:24 48
EL
yA y dA y ab ab
= =
∫
11
26
yb=
consent of McGraw–Hill Education.
PROBLEM 5.34
Determine by direct integration the centroid of the area shown.
Express your answer in terms of a and b.
SOLUTION
For y1 at
,xa=
22
2
2 , 2 , or b
y b b ka k a
= = =
Then
2
12
2b
yx
a
=
By observation,
2( 2) 2
bx
y xbb
aa
=− += −
Now
EL
xx=
and for
0,xa≤≤
22
11
22
12
and
2
EL
bb
y y x dA y dx x dx
aa
= = = =
For
2,ax a≤≤
22
12 and 2
22
EL
bx x
y y dA y dx b dx
aa
==−==−
Then
2
2
2
0
2
2
3
200
22
27
2
32 6
aa
a
a
a
bx
A dA x dx b dx
a
a
bx a x
b ab
a
a
== +−
= +− − =
∫∫ ∫
and
2
2
2
0
2
43
2
200
2 22 2 3
2
22
2
43
11
(2 ) ( ) (2 ) ( )
23
7
6
aa
EL a
aa
bx
x dA x x dx x b dx
a
a
bx x
bx a
a
abbaa aa
a
ab
= +−
= +−
=+ −+ −
=
∫∫ ∫
consent of McGraw–Hill Education.
SOLUTION Continued
2
22
22
00
2
3
25 2
40
2
222
2
22
5 23
17
30
aa
EL
a
a
a
b b bx x
y dA x x dx b dx
aa
aa
bx b a x
a
a
ab
= +−−
= +− −
=
∫∫ ∫
Hence,
2
77
:66
EL
xA x dA x ab a b
= =
∫
xa
=
2
7 17
:6 30
EL
yA y dA y ab ab
= =
∫
17
35
yb=
PROBLEM 5.35
Determine the centroid of the area shown when a = 4 in.
SOLUTION
We have
111
1
22
EL
EL
xx
yy x
=
= = −
and
1
1
dA ydx dx
x
= = −
Then
( ) ( )
1
1
11
00
ln (1 2ln )
a
aa
a
a
a
dx
A dA adx ax x a
x
== += + =+
∫∫ ∫
and
[ ]
1
122
1
1
00
1 12 1
22 2
a
aa
a
EL a
a
dx ax a
x dA xdx x x a
x aaa
−
= + = + = +− =
∫ ∫∫
[ ]
122
1
22
120
00 1
1 1 11 1 2 1
2 2 2 2 22 2 2
2
a
aa
aa
EL
aa
dx a a a a
y dA y dx a dx x x aa
x
−
= = + = − =− +=
∫∫ ∫∫
Because of symmetry, computation of only one coordinate was necessary.
( )
( )
2
2
21
: 1 2ln 2
21
2 1 2ln
EL
a
xA x dA x a a
a
xy aa
−
= +=
−
= = +
∫
Find
x
and
y
when
4 in.a=
2
2(4) 1
PROBLEM 5.36
Determine the centroid of the area shown in terms of a.
SOLUTION
PROBLEM 5.31
Determine by direct integration the centroid of the area shown.
SOLUTION
For the element (EL) shown,
22
b
y ax
a
= −
and
()
()
22
22
()
1()
2
2
EL
EL
dA b y dx
ba a x dx
a
xx
y yb
b
a ax
a
= −
=−−
=
= +
= +−
Then
()
22
0
a
b
A dA a a x dx
a
== −−
∫∫
To integrate, let
22
sin : cos , cosxa a x a dxa d
θ θ θθ
= −= =
Then
/2
0
/2
22
0
( cos )( cos )
2
sin sin
24
14
b
A aa a d
a
baa
a
ab
π
π
θ θθ
θθ
θ
π
= −
= −+
= −
∫
and
()
22
0
/2
2 2 2 3/2
0
3
1()
23
1
6
a
EL
b
x dA x a a x dx
a
ba
x ax
a
ab
π
= −−
= +−
=
∫∫
SOLUTION Continued
()()
22 22
0
2 23
2
22
00
2
2
( ) 3
22
1
6
a
EL
a
a
bb
y dA a a x a a x dx
aa
b bx
x dx
aa
ab
= +− −−
= =
=
∫∫
∫
2
1
:1
46
EL
xA x dA x ab a b
π
= −=
∫
2
or 3(4 )
a
x
π
=−
2
1
:1
46
EL
yA y dA y ab ab
π
= −=
∫
2
or 3(4 )
b
y
π
=−
PROBLEM 5.32
Determine by direct integration the centroid of the area shown.
SOLUTION
2
21
,
x
y y kx
k
= =
But
2,a ka=
thus,
2
1
ka
=
2
21
,x
y ax y a
= =
2
21
2
0
3
3/2 2
0
()
21
3 33
EL
a
a
xx
x
dA y y dx ax dx
a
x
A dA ax dx
a
x
ax a
a
=
=−=−
= = −
= −=
∫∫
2 34
3/2 5/2 3
00 0
23
5 4 20
a
aa
EL
x xx
x dA x ax dx ax dx ax a
aaa
= − = − = −=
∫∫ ∫
23
13 9
:3 20 20
EL
a
xA x dA x a a x
= = =
∫
By symmetry,
9
20
a
yx= =
consent of McGraw–Hill Education.
PROBLEM 5.33
Determine by direct integration the centroid of the area shown. Express
your answer in terms of a and b.
SOLUTION
For y2 at
,xa=
22
, , or a
y b a kb k b
= = =
Then
1/2
2
b
yx
a
=
Now
EL
xx=
and for
0,
2
a
x≤≤
1/2
2
1/2
2
22
EL
ybx
ya
x
dA y dx b dx
a
= =
= =
For
,
2
axa
≤≤
1/2
12
11
()
2 22
EL bx x
y yy aa
= + = −+
1/2
21
1
() 2
xx
dA y y dx b dx
a
a
= − = −+
Then
1/2 1/2
/2
0 /2
1
2
aa
a
x xx
A dA b dx b dx
a
aa
= = + −+
∫∫ ∫
/2 3/2 2
3/2
0/2
3/2 3/2
3/2
2
2
2 21
3 3 22
2()
32 2
11
( ) ()
2 22 2
13
24
a
a
a
b xx
xb x
a
aa
ba a
a
a
aa
ba a
a
ab
= + −+
= +−
+− − + −
=
consent of McGraw–Hill Education.
SOLUTION Continued
and
1/2 1/2
/2
0 /2
1
2
aa
EL a
x xx
x dA x b dx x b dx
a
aa
= + −+
∫∫ ∫
/2 5/2 3 4
5/2
0/2
5/2 5/2
5/2
32
32
2
22
5 5 34
2()
52 2
11
() ()
3 24 2
71
240
a
a
a
b x xx
xb a
aa
ba a
a
a
aa
ba a
a
ab
= + −+
= +−
+− − + −
=
1/2 1/2
/2
0
1/2 1/2
/2
/2 3
2 22
2
0/2
22
2
2
2
11
22 2
1 11
22 2 2 3 2
1
()
4 2 2 6 22
a
EL
a
a
a
a
a
bx x
y dA b dx
aa
bx x x x
b dx
aa
aa
b bx x
x
a a aa
b a a ba
a
aa
=
+ −+ −+
= + −−
= +− − −
∫∫
∫
3
2
11
48
ab
=
Hence,
2
13 71
:24 240
EL
xA x dA x ab a b
= =
∫
17
130
xa=
2
13 11
:24 48
EL
yA y dA y ab ab
= =
∫
11
26
yb=
consent of McGraw–Hill Education.
PROBLEM 5.34
Determine by direct integration the centroid of the area shown.
Express your answer in terms of a and b.
SOLUTION
For y1 at
,xa=
22
2
2 , 2 , or b
y b b ka k a
= = =
Then
2
12
2b
yx
a
=
By observation,
2( 2) 2
bx
y xbb
aa
=− += −
Now
EL
xx=
and for
0,xa≤≤
22
11
22
12
and
2
EL
bb
y y x dA y dx x dx
aa
= = = =
For
2,ax a≤≤
22
12 and 2
22
EL
bx x
y y dA y dx b dx
aa
==−==−
Then
2
2
2
0
2
2
3
200
22
27
2
32 6
aa
a
a
a
bx
A dA x dx b dx
a
a
bx a x
b ab
a
a
== +−
= +− − =
∫∫ ∫
and
2
2
2
0
2
43
2
200
2 22 2 3
2
22
2
43
11
(2 ) ( ) (2 ) ( )
23
7
6
aa
EL a
aa
bx
x dA x x dx x b dx
a
a
bx x
bx a
a
abbaa aa
a
ab
= +−
= +−
=+ −+ −
=
∫∫ ∫
consent of McGraw–Hill Education.
SOLUTION Continued
2
22
22
00
2
3
25 2
40
2
222
2
22
5 23
17
30
aa
EL
a
a
a
b b bx x
y dA x x dx b dx
aa
aa
bx b a x
a
a
ab
= +−−
= +− −
=
∫∫ ∫
Hence,
2
77
:66
EL
xA x dA x ab a b
= =
∫
xa
=
2
7 17
:6 30
EL
yA y dA y ab ab
= =
∫
17
35
yb=
PROBLEM 5.35
Determine the centroid of the area shown when a = 4 in.
SOLUTION
We have
111
1
22
EL
EL
xx
yy x
=
= = −
and
1
1
dA ydx dx
x
= = −
Then
( ) ( )
1
1
11
00
ln (1 2ln )
a
aa
a
a
a
dx
A dA adx ax x a
x
== += + =+
∫∫ ∫
and
[ ]
1
122
1
1
00
1 12 1
22 2
a
aa
a
EL a
a
dx ax a
x dA xdx x x a
x aaa
−
= + = + = +− =
∫ ∫∫
[ ]
122
1
22
120
00 1
1 1 11 1 2 1
2 2 2 2 22 2 2
2
a
aa
aa
EL
aa
dx a a a a
y dA y dx a dx x x aa
x
−
= = + = − =− +=
∫∫ ∫∫
Because of symmetry, computation of only one coordinate was necessary.
( )
( )
2
2
21
: 1 2ln 2
21
2 1 2ln
EL
a
xA x dA x a a
a
xy aa
−
= +=
−
= = +
∫
Find
x
and
y
when
4 in.a=
2
2(4) 1
PROBLEM 5.36
Determine the centroid of the area shown in terms of a.
SOLUTION