area bounded by arc and two lines

area bounded by arc and two lines

Let

r

=

r

⁢

(

φ

)

be the equation of a

continuous

curve in

polar coordinates

and

A

be the area of the planar region by the curve and the

line segments

from the origin to two points of the curve corresponding the

polar angles

α

and

β

(

>

α

).  Then the area can be calculated from

A

=

1

2

⁢

∫

α

β

[

r

⁢

(

φ

)

]

2

⁢

𝑑

φ

.

(1)

Proof.

We fit between

α

and

β

a set of values

φ

1

<

φ

2

<

…

<

φ

n

-

1

(2)

and denote

α

=

φ

0

,

β

=

φ

n

and think the line segments from the origin to each point of the curve corresponding the values

φ

i

.  Then the region is divided into

n

parts.  For every part we form

inscribed

and

circumscribed

circular sector

with the common tip in the origin and the radii along the lines

φ

=

φ

i

.  The union of the inscribed sectors is contained in the region and the union of the circumscribed sectors contains the region.  The unions have the areas

∑

i

=

1

n

1

2

⁢

r

i

2

⁢

(

φ

i

-

φ

i

-

1

)

and

∑

i

=

1

n

1

2

⁢

R

i

2

⁢

(

φ

i

-

φ

i

-

1

)

,

where

r

i

means the least and

R

i

the greatest value of

r

⁢

(

φ

)

on the

interval

[

φ

i

-

1

,

φ

i

]

.  Hence the area

A

is between these sums for any

division

of the interval

[

α

,

β

]

with the values of (2).  But by the

definition

of the

Riemann integral

we know that there is only one

real number

having this

property

for any division and that also the

definite integral

∫

α

β

1

2

⁢

[

r

⁢

(

φ

)

]

2

⁢

𝑑

φ

=

1

2

⁢

∫

α

β

[

r

⁢

(

φ

)

]

2

⁢

𝑑

φ

is between those sums.  Q.E.D.

Example 1.

Determine the area

A

enclosed by the

lemniscate of Bernoulli

r

=

cos

⁡

2

⁢

φ

.

The portion of the

lemniscate

situated in the first

quadrant

is gotten when

φ

gets the values from 0 to

π

4

, whence we have

A

4

=

1

2

⁢

∫

0

π

4

(

a

⁢

cos

⁡

2

⁢

φ

)

2

⁢

𝑑

φ

=

a

2

2

⁢

∫

0

π

4

cos

⁡

2

⁢

φ

⁢

d

⁢

φ

=

a

2

2

⁢

/

0

π

4

⁡

sin

⁡

2

⁢

φ

2

=

a

2

4

and therefore the whole area in question is

a

2

.

Example 2.

Determine the area

A

enclosed by the

logarithmic spiral

r

=

C

⁢

e

k

⁢

φ

and two radii

r

1

:=

C

⁢

e

k

⁢

φ

1

and

r

2

:=

C

⁢

e

k

⁢

φ

2

(

k

>

0

,

φ

1

<

φ

2

).

The (1) directly yields

A

=

C

2

2

⁢

∫

φ

1

φ

2

e

2

⁢

k

⁢

φ

⁢

𝑑

φ

=

C

2

2

⁢

/

φ

=

φ

1

φ

2

⁡

e

2

⁢

k

⁢

φ

2

⁢

k

=

C

2

4

⁢

k

⁢

(

e

2

⁢

k

⁢

φ

2

-

e

2

⁢

k

⁢

φ

1

)

=

r

2

2

-

r

1

2

4

⁢

k

.

References

1

Ernst Lindelöf

:

Johdatus korkeampaan analyysiin

. Fourth edition. Werner Söderström Osakeyhtiö, Porvoo ja Helsinki (1956).

2

N. Piskunov:

Diferentsiaal- ja integraalarvutus kõrgematele tehnilistele õppeasutustele

.  Kirjastus Valgus, Tallinn (1966).

Title

area bounded by arc and two lines

Canonical

name

AreaBoundedByArcAndTwoLines

Date of creation

2013-03-22 19:05:15

Last modified on

2013-03-22 19:05:15

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

15

Author

pahio (2872)

Entry type

Derivation

Classification

msc 51-01

Classification

msc 53A04

Synonym

area in polar coordinates

Related topic

SectorOfACircle

Related topic

AreaOfPlaneRegion

Related topic

SubstitutionNotation

Generated on Fri Feb 9 21:55:01 2018 by

LaTeXML