James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

670 Chapter 10 Parametric Equations and Polar Coordinates
It therefore appears plausible (and can in fact be proved) that the formula for the area A
of the polar region 5 is
3
A − y b
a
1
2 f f sdg2 d
Formula 3 is often written as
4
A − y b
a
1
2 r 2 d
with the understanding that r − f sd. Note the similarity between Formulas 1 and 4.
When we apply Formula 3 or 4, it is helpful to think of the area as being swept out by
a rotating ray through O that starts with angle a and ends with angle b.
Example 1 Find the area enclosed by one loop of the four-leaved rose r − cos 2.
r=cos 2¨ ¨= π 4
SOLUTION The curve r − cos 2 was sketched in Example 10.3.8. Notice from Figure
4 that the region enclosed by the right loop is swept out by a ray that rotates from
− 2y4 to − y4. Therefore Formula 4 gives
FIGURE 4
¨= 5π 6
¨=_ π 4
r=3 sin ¨
¨= π 6
A − y y4
2y4
− y y4
0
1
2 r 2 d − 1 2 y y4
cos 2 2 d − y y4
cos 2 2 d
2y4
0
1
2 s1 1 cos 4d d − 1 2 f 1 1 4 sin 4g y4
0 −
8
Example 2 Find the area of the region that lies inside the circle r − 3 sin and outside
the cardioid r − 1 1 sin .
SOLUTION The cardioid (see Example 10.3.7) and the circle are sketched in Figure
5 and the desired region is shaded. The values of a and b in Formula 4 are determined
by finding the points of intersection of the two curves. They intersect when
3 sin − 1 1 sin , which gives sin − 1 2 , so − y6, 5y6. The desired area can be
found by subtracting the area inside the cardioid between − y6 and − 5y6 from
the area inside the circle from y6 to 5y6. Thus
n
O
r=1+sin ¨
A − 1 2 y 5y6
s3 sin d 2 d 2 1 2 y 5y6
s1 1 sin d 2 d
y6
y6
FIGURE 5
Since the region is symmetric about the vertical axis − y2, we can write
s1 1 2 sin 1 sin 2 d dG
A − 2F1
2 y y2
9 sin 2 d 2 1 2 y y2
y6
y6
− y y2
s8 sin 2 2 1 2 2 sin d d
y6
− y y2
s3 2 4 cos 2 2 2 sin d d
y6
fbecause sin 2 − 1 2 s1 2 cos 2dg
y2
− 3 2 2 sin 2 1 2 cos g y6 −
n
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