APPLICATION OF APPLICATION OF INTEGRATIONINTEGRATION

The integration can be used to The integration can be used to determine the area bounded by the determine the area bounded by the plane curves, arc lengths volume and plane curves, arc lengths volume and surface area of a region bounded by surface area of a region bounded by revolving a curve about a line.revolving a curve about a line.

I. AREA OF THE PLANEI. AREA OF THE PLANE REGIONREGION

We know that the area bounded by We know that the area bounded by a Cartesian curve y = f(x), x – axis, a Cartesian curve y = f(x), x – axis, between lines x = a & x = b between lines x = a & x = b given by given by

b

a

Area f (x)dx

b

a

Area f (x)dx

xx

YY

(0, 0)(0, 0)

r=f()

X

= c1

= c2

0

Y

Area=(1/2)r2

o

r

Example 1:Example 1: To find the area lying between the To find the area lying between the

parabola y =4x – xparabola y =4x – x22 and the line y = x. and the line y = x. The required area = ( The area The required area = ( The area

bounded by the parabola y =4x – xbounded by the parabola y =4x – x22, , the x – axis, in between lines x = 0 and the x – axis, in between lines x = 0 and x = 3) – the area bounded by y = x, x – x = 3) – the area bounded by y = x, x – axis, in between lines x = 0 and x = 3.axis, in between lines x = 0 and x = 3.

y = 4x-x 2

y = x

Y

X(0,0)

(3,3)

x = 3

3 3

2

x 0 0

9Area 4x x dx xdx2

Example 2 :Example 2 : To Sketch and find the area bounded by To Sketch and find the area bounded by

the loop of the curve 3aythe loop of the curve 3ay22 = x( x – a) = x( x – a)2.2.

The curve is symmetric about the x – The curve is symmetric about the x – axis, axis,

Therefore y is defined if x Therefore y is defined if x 0.0.

22 x x ax(x a)y y

3a 3a

The curve intersects x – axis at The curve intersects x – axis at (0,0) & (a, 0) (0,0) & (a, 0)

Therefore loop in formed between Therefore loop in formed between these point.these point.

x = 0 i.e., y – axis is tangent to x = 0 i.e., y – axis is tangent to the curve at the origin. the curve at the origin.

The curve does not have any The curve does not have any asymptotes. asymptotes.

Graph of the given curve

X

Y

(a,o)

0

The area bounded by the loop of the The area bounded by the loop of the curve curve

= 2 area bounded by the portion of = 2 area bounded by the portion of the loop of the curve, x – axis lying in the loop of the curve, x – axis lying in the first quadrant the first quadrant

a a

0 0

2

x (x a)2 ydx 2 dx3a

8 a15 3

Example 3:Example 3: To find the area inside the cardioideTo find the area inside the cardioide r = a ( 1+ cosr = a ( 1+ cos) and the circle r = 2) and the circle r = 2 cos cos Required area is given byRequired area is given by

XX

/ 2=

X

AreaArea

222

0

2 2 2

0

2 22 2 2

0 0 0

2 2

22

1 a2 a 1 cos d2 2

a 1 cos 2cos d a

a 2 1d 2 cos d cos d a

1a 2 2 . 02 2 2

aa2 2

Example 4:Example 4: To find the area bounded by the curve To find the area bounded by the curve yy22(a-x) = x(a-x) = x3 3 and its asymptote.and its asymptote. X = a is the asymptote to the curve.X = a is the asymptote to the curve. The required area is given by The required area is given by Area = 2 The area bounded by the Area = 2 The area bounded by the

curve and the asymptote lying in the curve and the asymptote lying in the first quadrant. first quadrant.

X

a

0a

02

222 4 2

0

2 ydx

x2 x dxa x

Put x = a sin

3 1 3 a4 a sin 4a . .4 2 2 4

Y

oX=a

EXERCISESEXERCISES 1. Find the area bounded by the one arch 1. Find the area bounded by the one arch

of the cycloid x = a (of the cycloid x = a ( - sin - sin ) ,) , y = a ( 1- cos y = a ( 1- cos ) and its base) and its base 2. Find the area of the region lying above 2. Find the area of the region lying above

x – axis, included between the circle xx – axis, included between the circle x22 + + yy22 = 2ax and the parabola y = 2ax and the parabola y22 = ax. = ax.

3. Find the area between the curve 3. Find the area between the curve x ( xx ( x22 + y + y22) = a(x) = a(x22 – y – y22) and its ) and its

asymptote.asymptote.

4.Find the area bounded by the curve .4.Find the area bounded by the curve .

5.Find the area bounded by the loops of 5.Find the area bounded by the loops of the curve rthe curve r22 = a sin = a sin

6. Find the area inside r = a ( 1 – cos 6. Find the area inside r = a ( 1 – cos ) ) and outside r = a sin and outside r = a sin

7: Find the area common to the curves 7: Find the area common to the curves r = a ( 1+ cosr = a ( 1+ cos) and r = a ( 1 - cos) and r = a ( 1 - cos))

2 2 23 3 3x y a

8. Find the area inside r = a and 8. Find the area inside r = a and outside r = 2acosoutside r = 2acos..

9. Find the area bounded by the 9. Find the area bounded by the loops of the curve xloops of the curve x33 + y + y33 = = 3axy.3axy.

10. Find the area bounded by the 10. Find the area bounded by the loops of the curve r = a sin3loops of the curve r = a sin3..

RECTIFICATION-LENGTH OF THE RECTIFICATION-LENGTH OF THE PLANE CURVEPLANE CURVE

The rectification is the process of The rectification is the process of determining the length of the determining the length of the arch of a plane curve. We know arch of a plane curve. We know that the derivative of the arc of that the derivative of the arc of length of a curve is given bylength of a curve is given by

2

22

2 2

ds dyds dx 1 dx for a cartesian curve y = f(x).dx dx

ds drds d r d for a polar curve r = f( )d d

ds dx dyds d d for a parametric curved d d

x = f( )and y=g( )

The length of the arc of the curve is The length of the arc of the curve is given by given by

Arc length s = Arc length s = b

ads

2b

a

2b b

a a

2b b2

a a

dy 1 dx for a cartesian curve y = f(x)dx

dx dyds for a parametric = d d dd

curve x = f( ) & y g( )

ds dr d = r d for a polar cud d

rve r = f( )

To find the length of the arc of To find the length of the arc of the parabola xthe parabola x22 = 4ay measured = 4ay measured from the vertex to one extremity from the vertex to one extremity of the latus rectum.of the latus rectum.

Here Here 2

22 2

x dy xy 2a4a dx

ds dy 1Therefore 1 4a xdx dx 2a

The required arc lengthThe required arc length2a 2a

2 2

0 0

ds 1dx 4a X dxdx 2a

2a2

2 2 1

0

1 x 4a x4a x sinh a2a 2 2

a 2 log 1 2

To find the perimeter of the curve To find the perimeter of the curve

The parameter equation of this The parameter equation of this

curve is x = a coscurve is x = a cos33, y = a sin, y = a sin33..

2 2 23 3 3x y a

2 2ds dx dyd d d 3a cos sin

Therefore Perimeter of the curveTherefore Perimeter of the curve

=6a=6a

π2 2

0 0

ds4 4 3a sinθ cosθ dθd

To find the perimeter of the To find the perimeter of the curvecurve

r = a ( 1+ cosr = a ( 1+ cos).).

22

dr a sind

ds drrd d

2a cos 2

0

0

dsTherefore Perimeter of the cardioide =2 dd

2 2a cos d2

8a

EXERCISESEXERCISES

1.Find the length of the arc of one arch of 1.Find the length of the arc of one arch of the cycloid x = a(the cycloid x = a(+ sin+ sin), y = a ( 1- cos ), y = a ( 1- cos ).).

2.Find the length of the loop of the curve2.Find the length of the loop of the curve 3ay3ay22 = x (x – a) = x (x – a)22.. 3.Find the length of the arc of the curve of 3.Find the length of the arc of the curve of

the centenary y = c cosh(x/c) measured the centenary y = c cosh(x/c) measured from the vertex to any point (x ,y).from the vertex to any point (x ,y).

4.Find the length of arc of the loop of the 4.Find the length of arc of the loop of the curve r rcurve r r22 = a = a22 cos2 cos2..

VOLUME OF REVOLUTIONVOLUME OF REVOLUTION

Let a curve y = f(x) revolve about x–Let a curve y = f(x) revolve about x–axis. Then the volume of the solid axis. Then the volume of the solid bounded by revolving the curve y = bounded by revolving the curve y = f(x), in between the lines x = a and x f(x), in between the lines x = a and x = b, about x – axis is given by = b, about x – axis is given by

b2

a

volume y dx

If the curve revolves about y – axis, the If the curve revolves about y – axis, the volume is given by volume is given by

Examples :Examples : 1.To find the volume of the solid obtained 1.To find the volume of the solid obtained

by revolving one arch of the curve x = a by revolving one arch of the curve x = a ((+sin+sin) , y = a (1 + cos) , y = a (1 + cos) about its base ) about its base

b2

a

volume x dy

X- axis is the base of the curveX- axis is the base of the curve

Therefore the required volume Therefore the required volume

-a a

Y

Xo

0

a a2 2

a 0

2

0

2 2

0

3 3

02 3

y dx 2 y dx

dx2 y dd

2 a (1 cos ) a(1 cos )d

2 a (1 cos ) d

5 a

2. To find the volume bounded by 2. To find the volume bounded by revolving the curve yrevolving the curve y22(a – x ) = x(a – x ) = x33 about about its asymptote.its asymptote.

X = a is the asymptote to the curve. X = a is the asymptote to the curve. Shifting the origin to the point (a, 0), we Shifting the origin to the point (a, 0), we get the new coordinates X = x – a &get the new coordinates X = x – a &

Y = y – 0 = y.Y = y – 0 = y. Then the volume bounded by Then the volume bounded by

revolving the curve about the revolving the curve about the asymptote is given by asymptote is given by

2

y 0

2

y 0

a2

x 0a

23

2x 0

volume 2 x dy

2 x a dy

x x2 x a d dxa x

3a 2x2 x a x dx

2 a x

2Put x a sin ,x 0 0

x=a = 2

2 23 2 2 3 2 4

0 0

3

2 3

Volume

6a cos sin d 4a cos sin d

1 1 1 3 1a 6. . . 4. . . .4 2 2 6 4 2 2

a4

2.To find the volume of the solid 2.To find the volume of the solid bounded by revolving the cardioide r bounded by revolving the cardioide r = a(1+cos = a(1+cos ) about the initial line) about the initial line

Required volume Required volume 2a

2

x 0

y dx

where x = r cos = a(1-cos )cos y=r sin = a (1+ cos )sin .x = 2a = 0, x = 0 =

0

2 2 2

3 3 2

03

Therefore the volume

= a (1 cos ) sin d a(1 cos )cos

a sin (1 cos ) (1 2cos )d

8 a5

PROBLEMSPROBLEMS

The loop of the curve 3ayThe loop of the curve 3ay22=x(x – =x(x – a)a)22 moves about the x – axis ; find moves about the x – axis ; find the volume of the solid so the volume of the solid so generated.generated.

Find the volume of the spindle Find the volume of the spindle shaped solid generated by shaped solid generated by revolving the curve about x – axis revolving the curve about x – axis ..

SURFACE AREA OF REVOLUTIONSURFACE AREA OF REVOLUTION

The area of the surface of the solid obtained The area of the surface of the solid obtained by revolving about x – axis, the arc of the by revolving about x – axis, the arc of the curvecurve

y = f(x) intercepted between the points y = f(x) intercepted between the points whose abscissa are a and b , is given bywhose abscissa are a and b , is given by

b

a

dssurface area 2 y dxdx

Examples :Examples : 1.To find the area of the surface of 1.To find the area of the surface of

the solid generated by revolving on the solid generated by revolving on arch of the curve x= a ( arch of the curve x= a ( – sin – sin), y ), y = a(1 – cos= a(1 – cos) about its base.) about its base.

X =a(X =a( -sin -sin ) ,y = a(1 - cos ) ,y = a(1 - cos ))

2 2ds dx dy d d d

2a sin 2

2 y

02

02

02

Requried surface area = 2 yds

ds2 y dd

2 a(1 cos )2a sin d2

64 a3

2.To find the surface of the solid formed 2.To find the surface of the solid formed by revolving the curve r = a(1+cosby revolving the curve r = a(1+cos) ) about the initial line , about the initial line ,

22ds drr

d d

2a cos 2

0

0

2

dsTherefore surface area 2 y dd

2 r sin 2a cos d2

32 a5

EXERCISESEXERCISES

1.Find the area of the surface of the solid 1.Find the area of the surface of the solid generated by revolving the arc of the generated by revolving the arc of the parabola parabola

yy2 2 = 4ax bounded by its latus rectum about= 4ax bounded by its latus rectum about x –axis.x –axis. 2.Find the area of the surface of revolution 2.Find the area of the surface of revolution

about the x – axis the curve xabout the x – axis the curve x2/3 2/3 +y+y2/32/3 = a = a2/3 2/3 .. 3.Find the total area of surface of revolution 3.Find the total area of surface of revolution

of the curve rof the curve r22 = a = a22cos2cos2 about the initial line. about the initial line. 4.Find the area of the surface of revolution of 4.Find the area of the surface of revolution of

the loop of the curve 3aythe loop of the curve 3ay22=x(x-a)=x(x-a)22 about the about the x – axis.x – axis.