DDR 3 p.132-189

where limits and must be employed

x( ) a ,

y( ) b .

8.1.3. Area in polar coordinates

Let S be the region bounded

by

the

rays

1

α and 2 β and by the curve ρ ρ( ) , as

ρ ρ( )

shown in Fig. 2.11. The area of S is then given by

S 1

β

ρ2 ( ) d .

α β

2

α

О

8.2. Arc length

Fig. 2.11

arc

The length l in rectangular coordinates of the

y f (x) , x [a; b] , if f

(x)

is continuous in this interval, then

b

2

l

1

dx.

( f (x))

a

If the path is given parametrically,

x x(t) ,

y y(t)

and if x (t) ,

y (t)

are continuous in the interval

[t1 ; t2 ] , then arc length is found according to

the following formula

t2

2

2

l

dt,

(x (t))

( y (t))

t1

2

2 2 d

l

1

y

y

y=f(x)

S(x)

О a

b x

О a

x

b x

Fig. 2.12

Fig. 2.13

b

V f 2 (x)dx.

a

Let y = f1(x)

and y = f2(x) be two continuous functions such that

0 f1 (x) f 2 (x)

for x in the interval [a, b]. Let S be the region bounded by

2

2

dt.

P 2 y(t) (x (t))

( y (t))

b

A F(x)dx.

a

8.6. The centroid

The centroid of a curve l, such that has an equation

y f (x) ,

x [a, b]

is defined as the point (xc, yc), where

b

b

2

b

b

2

xdl

x 1

dx

f (x)dl

dx

( f (x))

f (x) 1 ( f (x))

x

a

a

,

y

a

a

.

b

b

b

b

c

2

c

2

dl

dx

dl

dx

1 ( f (x))

1 ( f (x))

a

a

a

a

b

1

b

xf (x)dx

f 2 (x)dx

x

a

,

y

2

a

.

b

b

c

c

f (x)dx

f (x)dx

a

a

Typical problems

T8.

Area

y x 2 3x and the

1. Find the area of the region bounded by the curve

and y = 4. We obtain

y x2 3x

x1 1; x2 4;

y 4

y1

4; y2

4.

4

4

S [4 (x 2 3x)]dx [4 x 2 3x]dx

1

1

y

æ

x

3

3

ö

4

64

4

ç

2

÷

=ç4x -

+

x

÷

= 16

24

ç

÷

3

ç

3

2

÷

ø

-1

è

( 4

1

3

)

20

5 .

3 4 x

3

2

6

–1

2. Find the area of the region bounded by the

Fig. 2.14

curve y

x and the lines

y x 2 ,

y x

( see

Fig. 2.15 ).

Solution. The line

x 1 is passed through the intersection of the straight

lines y x 2 and

y x ,

and is

divids the

region

by

two

parts. Then

S S1 S2 , where

S

is the given region,

S1

is the left part and is bounded

by the lines

y x ,

x 1 and the curve

y

x , S2

is the right part of S

and is bounded by the lines y x 2 ,

x 1 and

y

x . We have:

1

3

1

2

1

5

2

S1 (

2x

x 2

x x)dx

;

3

3

2

6

0

2

0

4

3

4

x 2

2x

2

16

2

1

19

S2 ( x x 2)dx

2x

8

8 (

2)

;

3

2

3

3

2

6

1

1

S S1 S2

5 19

4 .

6

6

3.

Find the

area

of the region

bounded

by

the

two parabolas

x y 2

2y 3

and x 5 4 y y 2 .

Solution.

Transforming

the

equations we

get

x 4 ( y 1)2

and

(x 9) ( y 2)2 .

у

х=1

у=х–

у

4

В

х=5+4у–у2

О

2

4

х

–4

9 х

5

–

у=–х

–1

2

Fig. 2.15

Fig. 2.16

The

region

is sketched

in

Fig.

2.16. The curves

intersect

when

y 2 2y 3 5 4y y 2 . Thus y 2

3y 4 0 , or equivalently (y + 1)(y –4)

= 0, from which

y A 1 or

yB 4 . With vertical strips three integrals are

4

4

S [5 4y y 2

( y 2

2y 3)]dy [8 6y 2 y 2 ]dy

1

1

8y 3y 2

2 y3

4

32 48 128 ( 8 3

2

)

125 .

3

1

3

3

3

4. Find the area of the less region bounded by the ellipse x 6 cos t ,

y 4 sin t and the straight line

y 2 3 ( y 2 3 ) .

Solution. The region is sketched in Fig. 2.17. The given area S is equal to

S 2SMNC 2(SONCD SOMCD ) .

At the point N xN 0 , thereby cos t 0

and t N

. The point C is the

2

intersection of the line and ellipse, thereby 2

3 4 sin t

, sin t

3

, tC

.

2

3

SOMCD OM MC 2

3 6 cos

6 3 .

3

Evaluating the area of the region ONCD :

SONCD 3

4 sin td(6 cos t)

243 sin 2

tdt

242 sin 2 tdt

2

2

3

122

(1 cos 2t)dt 12(t sin 2t )

2

12(

3

) 2 3 3 .

2

2

3

4

3

3

Hence, S 2(2 3 3 6 3)

4 6 3 .

5. Find the area of the region bounded by the Bernoulli’s lemniscate

(x2 y 2 )2 x2

y 2 (Fig. 2.18).

Solution. Transform the given equation into a corresponding polar

equation. In this case the substitutions

x ρcos ,

y ρsin are

appropriated

4

2 (cos2 sin 2

) , thereby

cos 2 .

1

4

0

S

4S1

4

2

cos 2 d sin 2

1 .

4

0

y

B

N

C

у

4

M

S1

A

O

D

x

х

Fig.2.17

Fig. 2.18

π

= 3 ( ρ 3 ).

Solution.

The

curve

3

ρ=3

π

ρ 6 sin 3 is called

a three-

B

leaved

rose.

As increa-ses

6

π

C

from 0 up to

3 , 3 increases

18

О

A

from 0 up to . Thus , which is

sin 3 , goes from 0 up to 6, then

ρ=6

back to 0, for in [0,

]. This

3

gives one loop of the three loops

making

up

the

graph

of

Fig. 2.19

ρ 6 sin 3 . For 0

in [

,

2

],

3

3

ρ 6 sin 3 is negative

(or 0).

The second function = 3 is the circle. For

ρ 3 the given area S is

shaded in Fig. 2.21 and is equal to S 6S ABCA .

Finding the polar coordinates of А and

В:

6 sin 3 3 ,

sin 3

1

,

2

SOABO

1

6

36 sin 2

3 d 9 6

(1 cos 6 )d 9( sin 6 )

6

3

3 .

2

6

4

18

18

18

Solution. By the given formula l b

1 y 2 dx.

a