Lesson 12 centroid of an area

TOPIC
APPLICATIONS
CENTROIDS OF PLANE AREA

The mass of a physical body is a measure of the quantity of the matter in it,
whereas the volume of the body is a measure of the space it occupies.
If the mass per unit volume is the same throughout the body is said to
be homogeneous or to have constant density.
It is highly desirable in physics and mechanics to consider a given mass
as concentrated at a point, called its center of mass (also, its center of gravity).
For a homogeneous body, this point coincides with its geometric center or
centroid. For example, the center of mass of a homogeneous rubber ball
coincides with the centroid (center) of the ball considered as a geometric
solid (a sphere).
DISCUSSION
The centroid of a rectangular sheet of paper lies midway between the two
surfaces but it may well be considered as located on one of the surfaces
at the intersection of the two diagonals. Then the center of mass of a thin
sheet coincides with the centroid of the sheet considered as a plane area.

The Moment ML of a Plane Region with respect to a line L is the product
of the area and the directed distance of the centroid from the line. The
moment of a composite region with respect to a line is the sum of the
moments of the individual sub-regions with respect to the line.
The moment of a plane region with respect to a coordinate axis may
be found as follows:
3. Evaluate the definite integral of the product in step 2 and apply the
fundamental theorem.
1. Sketch the region; showing a representative strip.
2. Form the product of the area of the rectangle and the distance of its
centroid from the axis.

For a plane region having an area A, centroid and moments
and with respect to x and y axes,
( ),, yxC
xM yM
yM Ax= xM Ay=and
A
M
x
y
=
A
M
y x
=

Determine the centroid of the first-quadrant region bounded by the parabola
.
.4 2
xy −=
V(0,4)
y
x
dx
(0,2)
x x=
( ),C x y
Curve: 2
4y x= −
( )
( )
2
2
4
4
0,4
x y
x y
V
= − +
= − −
0y =
2
0 4 x= −
2
4
2
x
x
=
= ±
if
Solving for the area A:
∫ −=
2
0
2
)4( dxxA2
04
2
2
+−
=
+
=
xyy
y BA
y
04 2
−−=− xyy BA
dxxdA )4( 2
−=
EXAMPLE

0
2
3
4
3






−=
x
xA






−= )8(
3
1
)2(4
3
16
=A sq. units

( )
( )
( )
( )( )
( )
( ) ( ) ( )
2
2
0
2
2
0
2
2
0
2
2
2 2
0
2
2 4
0
2
3 5
0
4
2
4
2
1
4
2
sin 4
1
4 4
2
1
16 8
2
1
16 8
2 3 5
1 8 1
16 2 8 32
2 3 5
1 480 320 96
2 15
x
x
x
x
x
x
x
x
M A y
M x y dx
y
but y
y
M x dx
x y dx
ce y x
M x x dx
M x x dx
x x
M x
M
M
= ×
= − ×
=
= − ×
= − ×
= −
= − −
= − +
 
= − × + 
 
 
= − + 
 
− + 
=  
 
∫
∫
∫
∫
∫
1 256 128
.
2 15 15
xM cu units
 
= = 
 
Moment about the x-axis

Moment about the y-axis
( )
( )
( )
( ) ( )
2
2
0
2
2
0
2
3
0
2
2 4
0
2
2 4
0
4
4
4
4
2 4
1
2
4
1
2 4 16
4
8 4
4 .
y
y
y
y
y
y
y
y
y
M A x
M x x dx
but x x
M x xdx
M x x dx
x x
M
M x x
M
M
M cu units
= ×
= − ×
=
= −
= −
 
= × − 
 
 
= − 
 
= −
= −
=
∫
∫
∫

units
128
15
16
3
8
5
x
x
M A y
M
y
A
y
= ×
= =
= units
4
16
3
3
4
y
y
M A x
M
x
A
x
= ×
= =
=






5
8
,
4
3
:C

Determine the centroid of the fourth-quadrant area bounded by the curve
2
4y x x= −
.
V(2,-4)
dx
y
x
y−
2
y
y =
xx =
•
Curve: 2
4y x x= −
( )
( )
2
2
4 4 4
2 4
2, 4
x x y
x y
V
− + = +
− = +
−
( )
2
0; 4 0
4 0
0; 4
y x x
x x
x x
= − =
− =
= =
EXAMPLE

( )
4
0
2
4
2
0
4
4
dA y dx
A y dx
y x x
A x x dx
= − ×
= − ×
= −
= − −
∫
∫
but
Solving for area A:
( )
( ) ( )
4
2 3
4
2
0
0
4 4
2 3
1 64
2 16 64 32
3 3
96 64 32
.
3 3
or
x x
A x x dx
A
A sq units

= − = × − 

= − = −
−
= =
∫

( )
( )
( )
( ) ( ) ( )
4
0
4
0
2
2
0
2
4
2
0
4 2
2
0
4
4 3 2
0
4
5 4 3
0
2
2
1
4
2
sin 4
1
2
1
4
2
1
8 16
2
1
8 16
2 5 4 3
1 1 16
1024 2 256 64
2 5 3
x
x
x
x
x
x
x
x
M A y
M y y dx
y
but y
y
M y dx
x y dx
ce y x
M y dx
M x x dx
M x x x dx
x x x
M
M
= ×
= − × ×
=
= − × ×
= − ×
= −
= − ×
= − −
= − − +
 
= − − × + × 
 
 
= − − +
 
∫
∫
∫
∫
∫
∫
1 3072 7680 5120
2 15
1 512 256
.
2 15 15
x
x
M
M cu units

− + 
= −  
 
 
= − = − 
 
( )
( )
( )
( )
4
2
0
4
2
0
4
2 3
0
4
3 4
0
4
4
4
4
3 4
4
64 64
3
256
64
3
256 192
3
64
.
3
y
y
y
y
y
y
y
y
y
M A x
M x x x dx
but x x
M x x xdx
M x x dx
x x
M
M
M
M
M cu units
= ×
= − ×
=
= −
= −
 
= × − 
 
 
= − 
 
 
= − 
 
− 
=  
 
=
∫
∫
∫

_
x yAM =
3
32
15
256
A
M
y x
_
−
==
unitsy
5
8−
= units
64
3
32
3
2
y
y
M A x
M
x
A
x
= ×
= =
=





 −
∴
5
8
,2C

2
y x=
y x=
Determine the centroid of the region bounded by the curve
and the line .
.
dx
y
x
(1,1)
2
y x=
y x=
x x=
Curve: 2
y x=
( )0,0V
y x=Line:
Intersection point:
( )
2
2
0
1 0
0; 1
0; 0
1; 1
x x
x x
x x
x x
if x y
x y
=
− =
− =
= =
= =
= =
yA-yB
EXAMPLE

Solving for area A : dA = ydx
∫=
1
0
ydxA ∫ −=
1
0
CL dx)yy( but yL
= x, yC
= x2
∫ −=
1
0
2
)( dxxxA
0
1
3
x
2
x 32
−=
)1(
3
1
)1(
2
1
−=
6
1
A = square units

( )
( )
( )
( )
1
0
1
2 2
0
1 2
2 2
0
1
2 4
0
1
3 5
0
2
1
2
1
2
1
2
1
2 3 5
1 1 1
2 3 5
1 5 3
2 15
1
.
15
x
L C
x L C
x L C
x
x
x
x
x
M A y
y y
M y y dx
M y y dx
x x dx
M x x dx
x x
M
M
M
M cu units
= ×
+ 
= −  ÷
 
= −
 = −
  
= −
 
= − 
 
 
= − 
 
− 
=  
 
=
∫
∫
∫
∫
( )
( )
( )
units.cuM
M
xx
M
dxxxM
dxxxxM
dxxyyM
xAM
y
y
y
y
0
y
0
CLy
y
12
1
12
34
4
1
3
1
43
1
0
43
1
0
32
1
2
1
=
−
=−=



−=
−=
−=
−=
⋅=
∫
∫
∫

1
15
1
6
2
5
x
x
M A y
M
y
A
y units
= ×
= =
=
1
12
1
6
1
2
y
y
M A x
M
x
A
x units
= ×
= =
=
1 2
,
2 5
C
 
∴  ÷
 

2
x y= 2
8x y= −Determine the centroid of the area bounded by the parabolas and
.
2
x y=
( )0,0V
Curve 1:
dy
y
x
(x1,y) (x2,y)
(4,-2)
2
x y=
2
8x y= −
2 1x x−
x
( ),C x y
y
Curve 2: 2
8x y= −
( )0,0V
Solving for intersection points:
( )
( )
2
2
4
4
3
8
8
8 0
8 0
0; 2
0; 4
y y
y y
y y
y y
y y
x x
= −
= −
+ =
+ =
= = −
= =
Therefore, the intersection points
are (0, 0) and (4, -2).
EXAMPLE

( )
( )
( )
( )
( )
( ) ( ) ( )
[ ]
3
2
3
2
2 1
0
2 12
2
2
2
1
0
2
2
0
3
0
2
2
0
3
2
3
8
8
8
8
2
2 2
2 3
2 1
1
3 3
2
2 1
2 2 2
3 3
2 8
8
3 3
16 8
3 3
8
3
dA x x dy
A x x dy
x x y
x y
x y
x x y
A y y dy
y
A y dy
y
A y
A
A
A
A
−
−
−
−
−
= −
= −
= −
= ± −
= −
=
= − −

= − − − − 

−
= − −


 = − − + − 
− 
= +  
 
= −
=
∫
∫
∫
:
:
.sq units
Solving for area A:
Continue solving for the centroid

Find the centroid of each of the given plane region bounded by the following curves:
1. y = 10x – x2
, the x-axis and the lines x = 2 and x = 5
2. 2x + y = 6, the coordinate axes
3. y = 2x + 1, x + y = 7, x = 8
4. y2
= 2x, y = x – 4
5. y = x3
, y = 4x [first quadrant]
6. y2
= x3
, y = 2x
7. y = x2
– 4, y = 2x – x2
8. the first quadrant area of the circle x2
+y2
= a2
9. the region enclosed by b2
x2
+ a2
y2
= a2
b2
in the first
quadrant

Lesson 12 centroid of an area

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Lesson 12 centroid of an area