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3-Geometrical Applications of Integration

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200 Engineering Mathematics through Applications

Geometrical Applications of Integration

3

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3 INTRODUCTION

In general, we consider the integration as the inverse of differentiation. In the expression of

the sum, () ,

∑ fx x ∆ f is considered continuous on a ≤ x ≤ b and we find that limit of S as ∆ x

approaches to zero is the number ∫() =−() (),

fxdx Fb Fa where F is any anti-derivative of f. We

apply this contention in finding the area between the x -axis and the curve y = f ( x ), a ≤ x ≤ b. We

extend the application to compute distances, volumes and volumes of revolution, length of

curves, area of surface of revolution, average value of function, centre of mass, centroid, etc.

3 AREA OF BOUNDED REGIONS (QUADRATURE)

I. Areas of Cartesian Curves

The area bounded by the curve y = f ( x ), X -axis and the

ordinates x = a , x = b is ∫

ydx , when f ( x ) is continuous single

valued and finite function of x, and y does not change sign in the

interval [ a , b ].

If AB is the curve y = f ( x ) between the ordinates

LA ( x = a ) and MB ( x = b ) with a condition that y is strictly

increasing (or strictly decreasing) function of x in the

interval [ a , b ].

Let P ( x , y ) and Q ( x + δ x , y + δ y ) be two neighbouring

points on this continously increasing curve y = f ( x ) and

NP , N ’ Q be their respective ordinates.

Here clearly the ALNP i. A depends on the position of

the point P ( x , y ) whose abscissa is ‘ x ’ and area PN N ’ Q = δ A lies between the areas of the

rectangles PN ’ and NQ , i. δ A lies between area y δ x and ( y + δ y )δ x or

lies between y and

( y + δ y ).

On taking the limits as Q → P , i_._ δ x → 0 (meaning thereby δ y → 0), =

dA y dx

Integrating both sides between the limits x = a to x = b , we get

Fig. 3.

(, )xyP

P ́

y B

O L N N ́ Mx-axis

y = f(x)

x δx

Q ́

( + , + ) x x y y δ δ

yy y + δ

Geometrical Applications of Integration 201

Area

b b a a

ALMB == A ∫ ydx.

However, if x and y are interchanged in the above formula, we

see that the area bounded by the curve x = f ( y ), Y -axis and the abscissa

y = a , y = b is.

∫ xdy

Observations:

( i ) The area bounded by the curve y = f ( x ), the two ordinates at A and B and the X -axis is often called the

area under the curve AB and the process of calculating the area bounded by the curve is called quadrature.

( ii ) An area whose boundary is described in anti-clockwise direction is considered positive otherwise

negative. Or in other words, for the portion of the curve (under

consideration) above X -axis for which y is positive, area enclosed is

positive, whereas for the portion of the curve (under consideration)

below X -axis for which y is negative, area is negative_._

But in case of area with negative sign, we mean numerical value of

the area.

( iii ) If in the interval, a ≤ x ≤ c , the curve y = f ( x ) is above the x -axis and in the

interval c ≤ x ≤ b , the curve y = f ( x ) is below the x -axis then we write the

area

∫∫∫=+

bc b

aac

ydx ydx ydx

or in otherwords, if x -changes sign from a to b , y = f ( x ) changes sign at

some interval point x = c (say), then the area for x from a to c and

c to b , are calculated seperately and then their numerical value

are added (see Fig. 3)

Similarly, the result can be extended if y changes sign at more

than one intermediate point in the interval [ a , b ].

( iv ) Area of the region bounded between two continuous curves f ( x )

and g ( x ) on [ a , b ] and the vertical lines x = a , x = b is given by

=− ∫ [() ()]

Afxgxdx ,

where f ( x ) ≥ g ( x ) in [ a , b ]. (see Fig. 3)

In the region under consideration, representative rectangle is

shown with height: f ( x i ) – g ( x i ), width: ∆ x and P ( x i g ( x i )); Q ( x i f ( x i ))

If f ( x ) ≥ g ( x ) in [ a , c ] and f ( x ) ≤ g ( x ) in [ c , b ], then we write the area as

=− +−∫∫[() ()] [() ()]

Afxgxdxgxfxdx

as shown in Fig. 3.

gx()i

∆x

f()xi

fx()

gx()

O a x i b

xa = xc = xb =

fx() gx()

Fig. 3 Fig. 3.

L M X O

xa = xb =

yfx = ( )

Fig. 3.

+ve area

xa =

O –ve area

xb =

xc = M X L N

Fig. 3.

M B

xfy = ( )

yb =

L ya = A

Geometrical Applications of Integration 203

2 2 22

cos 4

sin 4sin 4 cos

a aada a

π π

⋅θθ=

θ =⋅ θθ= θ

∫

Example 3:Find the area of the curve a

2 y

2 = x

3 (2 a – x ).

Solution: Without going into geometrical details of the curve, the area enclosed by it in the

first quadrant is

3 222

x ydx a x dx a

== −

∫∫

( i ) The curve intersects X -axis at x = 0, x = 2 a.

( ii ) Axis of X is the tangent at the origin.

( iii )

3 0at 2 , at 2

dy dy xa x a dx dx

== =∞=

Put x = 2 a sin

2 θ

3 22 2 2

(2 sin ) A 2 2 sin 4 sin cos

a aa a d a

=⋅

θ −θ⋅θθθ

∫

2242

16 sin ad cos

=θθθ

∫

⋅π π == ⋅⋅

2 2 31 16 6422 2

a a

( 1)( 3) ( 1)( 3) using sin cos ()( 2 ) 2

pp qq pqd pqpq

π −−...−−... π θθθ= ++− 

∫

Hence the total area is π a 2 .

Example 4: For the curve y

2 ( a – x ) = x

2 ( a + x ) [NIT Kurukshetra, 2004]

( i ) Find the area of the loop

( ii ) Area of the portion bounded by the curve and

its asymptote.

Solution: The curve passing through the origin is

symmetrical about X -axis and has x = a as its

asymptote. It intersects the axis of X at (– a , 0) as shown

in Fig. 3.

For the area of the loop, x varies from – a to 0.

Further, the loop is symmetrical about X -axis

∴

00 0

Area of the loop 2 2 2 aa a

ax ax ydx x dx x dx −− ax − ax

−−

∫∫ ∫

(on rationalization)

002

22 22

2 aa

dx dx

ax x

−− ax ax



 −−

∫∫

001 222 222 22

() 2() aa

dx dx

aax ax a x ax

−

−−

−− =−  − 

∫∫

X ́ X Aa(2 , 0)

O (0, 0) (, 0)a

  

3 , 2

Fig. 3.

x a=

= –

O A

(– , 0)a

Y ́

X ́

Fig. 3.

204 Engineering Mathematics through Applications

001 0 222 2 22 22

1 2()() aaa

adxa x xadx a xdx ax

−

−−−

−+

 =−− −−

−

∫∫∫

0 021 22 222 2 1 1 2() sin ins a 22 a

xxaxa x da x dx aa

−−

− −

−+

 − =−−+



∫

000 1222 222 1 2( ) sin 22 a a a

axxax ax a

−

− − −

−+

 − =−−  

()()()

2 21 200 sin(1) 00 , 2

a a

− +

 = −− − −−−  

where

− −π −=

1 sin ( 1) 2

π =−+ =⋅ −+π= π−  

22 2 2 2 2 [ 4 ] ( 4) numerically 44 2

aa a a

Alternately:

2, a

ax Ax dx − ax

= −

∫

Put x = a sinθ, dx = a cosθ d θ

Limits for x = 0, θ = 0; x = – a ,

−π θ= 2

(1 s i n ) 2sin osc (1 s i n )

a Aa ad a

−π

+θ =θ θθ −θ

∫

1sin 1sin 2sin osc 1sin 1sin

aad −π

+θ+θ =θ θθ −θ+θ

∫

00 222 2 22

sin cos (1 sin ) 2 [sin sin ] 1sin

adda −π −π

θ=

θθ+θ =θ+θθ −θ

∫∫

0 0 22

1sin 2sin( 1 cos2) os c 224

a da −π −π

θ=

θθ =θ+−θ−θ+−  

∫

=π−= −π

22 (4) (4)numerically 22

( b ) Area between the curve and its asymptote

== −

∫∫

aa ax ydx x dx ax

This integrand is same as in the case ( a ) simply limits are 0 to a.

12222 222 1

Area 2 ( ) sin ( 4) 222

a axxaxa aa x a

−

− =− − +− = π+



Example 5: Show that the area of the loop of the curve y 2 ( a + x ) = x 2 (3 a – x ) is equal to the

area between the curve and its asymptote. [KUK, 2002; NIT Kurukshetra, 2003]

206 Engineering Mathematics through Applications

Alternatively, y can also be expressed as

3 ax 4( ) aax yx x ax ax

− −+

and if ( a + x ) = 4a sin

2 θ

then

3 2 2222222 2 0 66

cos (4 sin ) 8 sin cos 8 (4 sin cos cos ) sin

ydx a a a d a a d

ππ

θ θ− ⋅ θ θ θ= θ θ− θ θ θ

∫∫ ∫

22222222

8 a (2 sin cos ) cos da 8 sin 2 cos d

ππ

=θθ−θθ= θ−θθ

∫∫

####### ()

2222

1cos4 1cos 4 cos 4 cos 2 22

a da d

ππ

−θ +θ −θ=−θ+θθ  

∫∫

ππ

=−

θθ θθ += +  

2 22

sin 4 sin 2 sin 4 sin 2 4 42 42

a a as at

, limit value is zero.

 =+=  

22 13 13 433 42 22

Example 6: Find the area included between the circle x

2 + y

2 = 2 ax and the parabola

y 2 = ax****. [NIT Kurukshetra, 2005]

Solution: The circle x

2 + y

2 = 2 ax is symmetrical

about X -axis, and passes through the origin with

centre as ( a , 0), radius ‘ a ’, Y -axis ( x = 0) as the

tangent at the origin. Further, x

2 + y

2 - 2 ax = 0

or ( x – a ) 2 + y 2 = a 2 means it intersects X -axis at

x – a = ± a or x = 0, 2 a.

In case of parabola, y

2 = ax , it intersects the circle

x 2 + y 2 - 2 ax = 0 at (0, 0), ( a , a ).

Hence the area outside the parabola and inside

the circle is covered under limits x = 0 and x = a.

(see geometry).

Hence the desired area = 2 Area OBCO

()()

2 of the circle of the parabola

=− yy dx

∫

ax x dx axdx

 =−−  

∫∫

III

222 2 222

00 0

I2 () cossin

aa ax x dx a a x dx a a a d

=−=−−=−θ⋅θθ

∫∫ ∫

Fig. 3.

yax 2 = y-axis

(, )aa

(, 0)a

x-axis (2 , 0)a

(– , )aa

Geometrical Applications of Integration 207

4 222

sin 4

a d

π θθ

∫

(Taking a – x = a cosθ)

3 2 2

2 II 33

a x ax dx a = a





  

∫

∴

π  π =−=−=π− 

42 2422 2( 3 8) 43 23 6

aa Aaa

Example 7: Find the area included between the parabola x

2 = 4 ay and the curve

a y xa

= +

Solution: The curve, x 2 = 4 ay represents an upward parabola symmetrical about Y -axis with

the origin as its vertex.

The curve

a y xa

= +

which is symme-

trical about Y -axis, does not pass through

the origin. Further, y = 0, i_._ axis of X is

an asymptote to it. It meets the Y -axis

at (0, 2 a ).

To find the points of intersection,

equating the two values of y , i_._

axa

= +

or x 4 + 4 a 2 x 2 - 32 a 4 = 0

or ( x

2 + 8 a

2 )( x

2 - 4 a

2 ) = 0

Rejecting ( x

2 + 8 a

2 ) = 0 (which gives imaginary values of ‘ x ’), we get

x 2 - 4 a 2 = 0, i. x = ± 2 a , and y = a.

Thus, these two curves intersects at (2 a , a ) and (–2 a , a ).

∴

232

22 0

8 The required area 2 2 44

a ax OABC OBC dx xa a

 =× = −

∫+

2 3 31

11 28 tan 2243

a xx a aaa

− =−  

()

π =−−−  

231 24 0 8 0 412

aa a

####### ()

 =π−= π− 

22 2 22 232 33

aa a

Example 8: Find the area between the curve x 3 + y 3 = 3 axy and its asymptote x + y + a = 0.

[KUK, 2001; NIT Kurukshetra, 2007]

(2 , )aa

B(0, 2 )a

xay 2 = 4

C (–2 , )aa

xa 22 + 4

8 a 3 y =

Fig. 3.

Geometrical Applications of Integration 209

2 3

4 (cos cos 2 ) 3

b d

=θ+θθ

∫

2 3 22

4sin2 3 sin 3 32 2

b ba

θ =θ+ ==  

Example 9: Show that area common to the two parabolas y 2 = 4 a ( x + a ) and

y 2 = 4 b ( b – x ) is abab +

8 () 3

Solution: Both the parabolas,

=+...  =−... 

4( ) ()

4( ) ( )

yaxa i

ybbx ii

are

symmetrical about X -axis. Parabola, y 2 = 4 a ( x + a ), has its

vertex at A (– a , 0) and latus ractum as 4 a , whereas the

parabola y 2 = 4 b ( b – x ), has its vertex at B ( b , 0) and latus

ractum as 4 b.

For intersection of these two, we get

4 a ( x + a ) = 4 b ( b – x )

or x ( a + b ) = b

2 - a

or x = ( b – a )

Hence the two parabolas intersects at L and M for

x = b – a.

∴ Area included between them = area ALBMA = 2 × area ALBNA

= 2(Area ALNA + Area LNBL )

−

−−

 =+ −  

∫∫

24 () 4 ()

ba b

aba

ax adx bb xdx

() 22 () 44 33

ba b

aba

xa bx ab

−

−−

 

 +− =+  −

   

331331 222222

888 8 00 () 333 3

ab b a ab ab a b ab

   =− −−=+=+

  

Example 10: Find the area common to the circle x 2 + y 2 = 4 and the ellipse x 2 + 4 y 2 = 9.

Solution: The equation x

2 + y

2 = 4 represents circle with centre (0, 0) and radius 2 units

where the equation x

2 + y

2 = 9 or +=





2 2

1 33

x y represents an ellipse with semi-major axis as 3

units and semi-minor axis as

units and for intersection of these two, we get

L yaxa

2 = 4 ( + )

2 = 4 b (b – x)

(– , 0)a A

O N Bb(, 0)

Fig. 3.

210 Engineering Mathematics through Applications

x 2 + 4(4 – x 2 ) = 9 or 3 x 2 = 7 i. =

Since the ellipse and the circle both are symmetrical about both the axis

∴ Required common area to the circle and the ellipse

= 4 × area common to them in the Ist quadrant

= 4[area OAPD ] = 4[Area OLPD + area LAPL ]

7 2 3 7 0 3

4 (value of ellipse) ydx yd (value of circle) x



=+  

∫∫



=−+−  

∫∫

7 2 322 7 0 3

1 49 4 2

xdx xdx

−

   =−+ +  

7 3 21

9 49 sin 223

xx x

−

  −+ 



2 21

7 3

4 4sin 222

xx x

77 11 1 7 77 7 2 9 9sin 0 0 4sin (1) 4 4sin 3 3 33 3 3 23

−− −   = −+ −++ − −−     

720 117 7 5 7 29 sin4 sin 4 33 27 2 3 3 12

−− π =+ +×−⋅−   

72 0 5 117 7 22 9sin4sin 33 3 27 12

−−

  = − +π+ −    

11 35 7 7 2 2 9sin 4sin 32712

−−

 = +π+ −   

ASSIGNMENT 1

1. Find the area of the ellipse

2 2

2. Find the area of the circle x 2 + y 2 = a 2

3. Find the area of the loop of the curve ay

2 = x

2 ( a – x )

HINT : xaxdx [] a ax axdx ( )

 −=−− −

∫∫

4. Calculate the area of the curve a

2 x

2 = y

3 ( a – y )

5. Find the whole area of the curve x 2 ( x 2 + y 2 ) = a 2 ( x 2 - y 2 ) [HINT : Put x 2 = a 2 cosθ]

xy + = 4

2 2

B P (0, 2)

xy + 4 = 9 2 2

D(0, 3/2)

O L

A(2, 0)

7 C(3, 0) 3

  

(2, 0)

Fig. 3.

212 Engineering Mathematics through Applications

= 2 a

2 (– π) + a

2 π = 3π a

2 .

Area between the curve and X -axis = 2 area OAL π == θ θ

∫∫

a dx ydx y d d

2(1cos)(1cos) aad

=−θ+θ∫ θ

/ 22 2 2 2

2 ada da sin 4 sin

ππ

=θ∫∫θ= θθ=π

∫ ⋅

−−... π θθ=  −... 

(1)(3) using sin (2) 2

pp d pp

Example 12: Show that the area of the hypocycloid x = a cos 3 t , y = b sin 3 t is

3 π ab

. Hence

deduce the area of the asteriod x = a cos 3 t , y = a sin 3 t****.

Solution: The given curve x = a cos 3 t , y = b sin 3 t (hypocycloid) shown in the figure, 3.

meets x -axis at t = 0 and t = π and to y -axis at

t and

t , and is symmetrical about

both the axis.

t = /2π

(0, )b

t = π

(– , 0)a t = 0

(, 0)a

(0, – )b

Fig. 3.

∴ The required area

a dx ydx y dt dt

==∫∫

0 32

4s btat t in.3cos dt

=−∫

/ 42

12 ab sin t cos t dt

=⋅∫

(4 1) (4 3) (2 1) 3 12 642 2 8

ab ab

−⋅ − ⋅ −ππ

⋅⋅ π  −−...−−...π =  ++−......

∫

(1)(3)1(1)(3) using sin cos ()( 2 ) 2

pq pp qq xxdx pqpq

Note : Area in case of asteriod

222 xy a 333 += which is a particular case of hypocycloid, when a and b are equal

becomes

π 2 3

a .

Geometrical Applications of Integration 213

Example 13 For any real t ,

tt tt ee ee xy

− − +− = , = 22

is a point

of the x 2 - y 2 = 1. Show that the area bounded by this

parabola and the lines joining its centre to the points

corresponding to t' and – t' is t'****.

Solution: Let P ( t' ) and Q (– t' ) be two points on the hyperbola

2 - y

2 = 1 (Fig. 3). Then the area bounded by the hyperbola

and the two lines OP and OQ is shown by the shaded portion.

The required area is the difference of the area of the ∆ OPQ

and area PAQBP.

Now

122 Area of ( ) 224

t' t' t' t' ee ee t' t' OPQ PB OB e e

−− +− − ∆=⋅= ⋅ = −

Area 2 2 2 22

Bt' t'tttt

dx ee ee PAQBP y dx y dt dt dt

−− −− == = ⋅

∫∫ ∫

− −  =+−=+−  −

∫

22 22

11 (2) 2 2222

t' t t t' tt ee ee dt t

− =−−

122 (4') 4

t' t' ee t

∴

22 22 11 The required area ( ) ( 4 ) 44

t' t' t' t' POQAP e e e e t t' −− =−−−−=

ASSIGNMENT 2

1. Show that the area of the loop of the curve }

sin 2 ,

sin

xa t

ya t

2. Show that the area bounded by the cissiod

=    =  

sin ,

sin

cos

xa t

t ya t

and its asymptote is π

3. Find the whole area of the curve

−  =  +   =   + 

t xa t

at y t

(It is the parametric form of the circle)

4. Find the area included between the cycloid

}

=−

(sin)

(1 c o s )

xat t

ya t

and its base.

Fig. 3.

Pt()

Qt ́()

xy + = 1 2 2

X O

OB x = =

ee tt'–' + 2

PB = =y

ee tt'–' - 2

Geometrical Applications of Integration 215

Example 14: Find the area of the loop of the curve r = a cos2 θθθθθ and hence find the total area

of the curve.

Solution: As we know that the curve r = a cos n θ or r = a sin n θ

have equal loops if n is odd and 2 n equal loops if n is even.

In our problem, r = a cos2θ, n is even means the curve has 4

equal loops.

Further to find the limits of integration for a loop, we

generally put r = 0 and find two consecutive values of θ.

Thus, there r = 0 implies cos2θ = 0 i_._

π 2 θ=± 2

or 4

π θ=±

i_._ for the first loop of the curve θ varies from

ππ to 44

as shown

in Fig. 3.

Area of one loop of the curve

/4 / 222

/4 /

1 1 cos 2 2 2

rd a d

ππ

−π −π

=θ= θθ

∫∫

2 / 22

2cos 2

a ad

=⋅ θθ

∫

As cos 2 is an even function of and for an even function

() 2 ()

fxdx fxdx −

θ θ 

 =  ∫∫

Putting 2θ = t , we get

Area

π ππ ===

∫

/2 2 2 22

1 cos 2222 8

dt a a at

The total area of the curve is 4 times the area of the single loop, i.

π ×

2 4 8

a or

Example 15: Find the area outside the circle r = 2 a cos θθθθθ and inside the cardiod r = a (1 + cos θθθθθ ).

[NIT Kurukshetra, 2008]

Solution: For the circle, r = 2 a cosθ,

θ= ⇒ =   π  θ= ⇒ = 

0 2

Further, the circle r = 2 a cosθ is symmetrical about the initial axis.

222

Otherwise also in cartisan coordinates, it becomes 2 cos or ( ) 2

( ) ( 0) , i. circle with centre ( , 0) and radius.

rr ar x y ax

xa y a a a

⋅=θ+=

  −+−= 

For cardiod, r = a (1 + cosθ);

0, 2 ;

,; 2

θ= = 

π  θ= = 

 θ=π = 

= 3 / 4 θπ = /

/ 4 θ

= 7 / 4 = – / 4

Fig. 3.

216 Engineering Mathematics through Applications

Further, the cardiod also is symmetrical about the initial line.

For intersection, the geometry is as shown in Fig. 3 (III)

Y-axis

O aa

θ = 0 X-axis Aa(2 , 0)

0, 2

π 

θπ = / (, /2)aπ

θπ = θ = 0

O (0, )π Aa(2 , 0)

θπ = /

θπ = θ

(2 , 0)a

dθ

(I) (II) (III)

Fig. 3.

Above the initial axis, the cardiod is traced from θ = 0 to π, whereas in case of circle, θ goes

from 0 to

So the area outside the circle inside the cardiod, i_._ Area OABDO.

= 2 [Area of the cardiod above X-axis – Area enclosed by circle above X-axis]

/ 22 12 00

11 2 22

rd rd

ππ  =θ−θ  

∫∫

/ 22 22

adad (1 co s ) 4 co s

ππ  =+θ− θθ  

∫∫

2 / 22 2

2cos 4 cos 2

add

ππ θ =θ−θθ  

∫∫

/ 24 2

4cos osc 2

ad d

ππ θ =θ−θθ  

∫∫

I II

24 /2 /2 2 00

4 cos cos In I, putting 2

atdt d t

ππθ  =−θθ = 

∫∫

2 2 31 1 4 422 22 2

a a

⋅π π π =−=  ⋅

(1)(3) using cos (2) 2

p pp d pp

π −−...π θθ=  −...

∫

Example 16: Show that the area included between the cardiod r = a (1 + cos θθθθθ ) and

r = a (1 – cos θθθθθ ) is

a π− 2 (3 8)

. [KUK, 2001]

218 Engineering Mathematics through Applications

Clearly the loop is bounded for r = 0, i. sinθcosθ = 0 or 0, 2

π θ=.

∴

/2 /2 22 22 332 00

11 sincos The required area of the loop 9 2 2 (sin cos )

rd a d

ππ θθ =θ= θ θ+ θ

∫∫

22 /2 2

32 0

9tansec

2( 1 tan)

a d

π θθ =θ +θ

∫

Put tan

3 θ = t , so that 3tan

2 θ sec

2 θ d θ = dt and limits are t = 0 to ∞

∞

∫

2 0

2(1)

adt

∞ ∞ − +  − ===  −+ 

221 2

33 (1 ) 13

21 21 2

aat a

For the curve r

2 = a

2 cos 2θ, one of the loop is bounded in between the radii vectors

−π θ= 4

, as r = 0 gives θ = ±π/4.

For details, see the Fig. 2 under tracing.

∴ Area of the loop

/4 /4 /4 2 22 2

000

1sin 2cos 222

a rd a d a

ππ π θ =θ= θθ= =  

∫∫

Whence the area of the loop of the curve x 3 + y 3 = 3 axy is three times the area of the one of

the loops of r

2 = a

2 cos 2θ.

Example 18: Find the ratio of the two parts into which the parabola 2 a = r (1 + cos θθθθθ )

divides the area of the cardiod r = 2 a (1 + cos θθθθθ ).

Solution: The curve r = 2 a (1 + cosθ) is a standard cardiod with values

θ = 0, r = 4 a , point A (4 a , 0)

π θ= 2

, r = 2, point B 2, 2

π  

θ = π, r = 0, point O (0, π)

π θ=

, r = 2 a , point C 2, 2

π − 

Likewise, for parabola =

+θ

1cos

a r ,

θ = 0, r = 2 a , point A ( a , 0)

π θ= 2

, r = 2 a , k point B

π



2, 2

π θ=

, r = 2 a , k point C 2, 2

π − 

Geometrical Applications of Integration 219

Clearly the two curves intersects at

±π θ= 2

, otherwise also,

==+θ +θ

2 2(1 cos ) 1cos

a ra ⇒ (1 + cosθ) 2 = 1 implying cosθ = 0, i_._

π θ=±

θπ = /

r a= 2 (1 + cos )θ Ba(2 , /2)π

(0, )π θπ = O θ = 0

Ca(2 , – /2)π

θπ = /

2 1cos

a r= +θ

Fig. 3.

222

11 The whole area of the cardiod 2 2 4 (1 cos ) 22

rd a d

ππ =θ= +θθ

∫∫

=4 ad (1 2cos cos )

∫+θ+θθ

222

1cos2 3 1 =4 1 2cos 4 2sin sin2 6 224

ada a

π π +θ  +θ+ θ= θ+θ+ θ=π  

∫

... (1)

Area of the unshaded region between the two curves = 2[Area OABO – area ODBO ]

ππ  =θ−θ  

∫∫

/2 / 22 21 00

11 2 22

rd rd ,

where r 1 = 2 a (1 + cosθ), cardiod and = +θ

1cos

a r , the parabola

/2 / 22 2 00

1 4( 1 cos) (1 c o s )

ad d

π π  =+θθ− θ  +θ 

∫∫

/2 / 2 2 00 2

3cos2 1 42 cos 22 2cos 2

ππ



 θ =+θ+θ−  θ

 

∫∫

/2 / 222

0 0

3sin2 1 4 2 sin sec sec 24422

ππ  θθθ =+θ+ − ⋅θ  

∫

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3-Geometrical Applications of Integration

Course: Kinematics of Machinery (19EARC202)

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Engineering Mathematics through Applications

200

Geometrical Applications of Integration

aaaaa

3.1 INTRODUCTION

In general, we consider the integration as the inverse of differentiation. In the expression of

the sum, () ,

afx x

∆

∑ f is considered continuous on a ≤ x ≤ b and we find that limit of S as ∆x

approaches to zero is the number =−

∫

() () (),

afxdx Fb Fa where F is any anti-derivative of f. We

apply this contention in finding the area between the x-axis and the curve y = f(x), a ≤ x ≤ b. We

extend the application to compute distances, volumes and volumes of revolution, length of

curves, area of surface of revolution, average value of function, centre of mass, centroid, etc.

3.2 AREA OF BOUNDED REGIONS (QUADRATURE)

I. Areas of Cartesian Curves

The area bounded by the curve y = f(x), X-axis and the

ordinates x = a, x = b is ∫

ydx, when f(x) is continuous single

valued and finite function of x, and y does not change sign in the

interval [a, b].

If AB is the curve y = f(x) between the ordinates

LA(x = a) and MB(x = b) with a condition that y is strictly

increasing (or strictly decreasing) function of x in the

interval [a, b].

Let P(x, y) and Q(x + δx, y + δy) be two neighbouring

points on this continously increasing curve y = f(x) and

NP, N ’ Q be their respective ordinates.

Here clearly the ALNP i.e. A depends on the position of

the point P(x, y) whose abscissa is ‘x’ and area PNN’Q = δA lies between the areas of the

rectangles PN’ and NQ, i.e. δA lies between area yδx and (y + δy)δx or A

δ lies between y and

(y + δy).

On taking the limits as Q → P, i.e. δx → 0 (meaning thereby δy → 0), =

dA y

Integrating both sides between the limits x = a to x = b, we get

Fig. 3.1

(, )

xyP

P´

-axis

OLNN´ M

yfx

= ()

Q´

( + , +)

xxyy

δδ

yy y

+ δ

3-Geometrical Applications of Integration

Kinematics of Machinery (19EARC202)

KLE University

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Geometrical Applications of Integration

3

∑ fx x ∆ f is considered continuous on a ≤ x ≤ b and we find that limit of S as ∆ x

approaches to zero is the number ∫() =−() (),

∫

∫∫

∫

∫

∫

∫∫ ∫

∫∫

∫∫

∫∫∫

∫

()()()

∫

∫

∫

∫∫

∫

∫∫

∫∫ ∫

∫∫

∫∫

()()

∫

∫∫

∫∫ ∫

∫

∫

∫+

()

∫

∫∫

∫∫

∫∫

HINT : xaxdx [] a ax axdx ( )

∫∫

∫∫

∫∫ ∫

∫

1. Show that the area of the loop of the curve }

}

∫∫

∫

∫

∫∫

∫∫

∫∫

∫∫

∫

∫∫

∫

∫

∫∫

∫∫

∫

∫∫

∫∫

∫∫

∫

3-Geometrical Applications of Integration

Course: Kinematics of Machinery (19EARC202)

University: KLE University

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