TOPIC

APPLICATIONSCENTROIDS 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,

,, yxCxM yM

yM Ax xM Ayand

A

Mx y

A

My x

Determine the centroid of the first-quadrant region bounded by the parabola .

.4 2xy

V(0,4)

y

x

dx

(0,2)

x x

,C x y

Curve: 24y x

2

2

4

4

0,4

x y

x y

V

0y 20 4 x

2 4

2

x

x

if

Solving for the area A:

2

0

2 )4( dxxA2

04

2

2

xyyy BA

y

04 2 xyy BA

dxxdA )4( 2

EXAMPLE

0

2

34

3

x

xA

)8(

3

1)2(4

3

16A sq. units

2 2

0

2 2

0

2 2

0

2

2 2 2

0

2 2 4

0

23 5

0

4

2

42

14

2

sin 4

14 4

21

16 82

116 8

2 3 5

1 8 116 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

ybut y

yM x dx

x y dx

ce y x

M x x dx

M x x dx

x xM x

M

M

1 256 128

.2 15 15xM cu units

Moment about the x-axis

Moment about the y-axis

2 2

0

2 2

0

2 3

0

22 4

0

22 4

0

4

4

4

42 4

12

4

12 4 16

48 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 xM

M x x

M

M

M cu units

units

12815163

8

5

x

x

M A y

My

A

y

units

4163

3

4

y

y

M A x

Mx

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)

dxy

x

y2

yy

xx

Curve: 2 4y x x

2

2

4 4 4

2 4

2, 4

x x y

x y

V

20; 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:

42 34 2

00

4 42 3

1 642 16 64 32

3 396 64 32

.3 3

or

x xA x x dx

A

A sq units

4

0

4

0

2 2

0

2

4 2

0

4 22

0

4 4 3 2

0

45 4 3

0

2

21

42

sin 4

1

21

421

8 162

18 16

2 5 4 3

1 1 161024 2 256 64

2 5 3

x

x

x

x

x

x

x

x

M A y

M y y dx

ybut y

yM y dx

x y dx

ce y x

M y dx

M x x dx

M x x x dx

x x xM

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

43 4

0

4

4

4

43 4

464 64

3

25664

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 xM

M

M

M

M cu units

_

x yAM

3

3215

256

A

My x_

unitsy5

8 units

643323

2

y

y

M A x

Mx

A

x

5

8,2C

2y xy x

Determine the centroid of the region bounded by the curve and the line .

.

dx

y

x

(1,1)

2y x

y x

x x

Curve: 2y x

0,0V

y xLine:

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

0CL 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

1A square units

1

0

1 2 2

0

1 22 2

0

1 2 4

0

13 5

0

2

1

21

21

2

1

2 3 5

1 1 1

2 3 5

1 5 3

2 15

1.

15

x

L Cx L C

x L C

x

x

x

x

x

M A y

y yM y y dx

M y y dx

x x dx

M x x dx

x xM

M

M

M cu units

units .cuM

M

xxM

dxxxM

dx xxxM

dx xyyM

xAM

y

y

y

y

0y

0 CLy

y

12

112

34

4

1

3

1

43

1

0

43

1

0

32

1 2

1

11516

2

5

x

x

M A y

My

A

y units

11216

1

2

y

y

M A x

Mx

A

x units

1 2,

2 5C

2x y 2 8x yDetermine the centroid of the area bounded by the parabolas and .

2x y

0,0V

Curve 1:

dy

y

x(x1,y) (x2,y)

(4,-2)

2x y

2 8x y

2 1x x

x

,C x y

y

Curve 2: 2 8x y

0,0VSolving for intersection points:

22

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

32

32

2 1

0

2 12

22

21

0 2

2

030

22

0

3

2

3

8

8

8

8

22 2

2 3

2 11

3 32

2 12 2 2

3 32 8

83 3

16 8

3 38

3

dA x x dy

A x x dy

x x y

x y

x y

x x y

A y y dy

yA y dy

yA 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 b2x2 + a2y2 = a2b2 in the first quadrant