Begin2.DVI

In Green’s theorem the functions M and N are arbitrary. Therefore, in the

special case M=−yand N= 0 one obtains

∫

C

©− y dx =

∫∫

R

dx dy =A=Area enclosed by C. (8 .25)

Similarly, in the special case M= 0 and N=x, the Green’s theorem becomes

∫

C

©x dy =

∫∫

R

dx dy =A=Area enclosed by C. (8 .26)

Adding the results from equations (8.25) and (8.26) produces

2 A=

∫

C

©x dy −y dx

or A=

1

2

∫

C

©x dy −y dx =

∫∫

R

dx dy =Area enclosed by C.

(8 .27)

Therefore the area enclosed by a simple closed curve Ccan be expressed as a line

integral around the boundary of the region R enclosed by C. That is, by knowing

the values of xand yon the boundary, one can calculate the area enclosed by the

boundary. This is the concept of a device called a planimeter , which is a mechanical

instrument used for measuring the area of a plane figure by moving a pointer around

the surrounding boundary curve.

Example 8-7.

Find the area under the cycloid defined by x=r(φ−sin φ), y =r(1 −cos φ) for

0 ≤φ≤ 2 πand r > 0 is a constant. Find the area illustrated by using a line integral

around the boundary of the area moving from A to B to C to A.

Solution

Let BCA and the line AB denote the bounding curves of

the area under the cycloid between φ= 0 and φ= 2π. The area

is given by the relation

A=

1

2

∫

C 1

(x dy −y dx ) +

1

2

∫

C 2

(x dy −y dx ),

where C 1 is the straight-line from Ato Band C 2 is the curve