SECTION 9.4
9.4
■
0 6
6 56
24. r 3 cos ,
6. r cos 3,
12 12
7. r 3 sin ,
4 34
9–16
■
■
■
■
■
■
■
■
■
■
■
27–32
■
14. r 1 sin 29. r 1 cos 15. r 3 cos 16. r sin 4
30. r e ,
■
■
■
■
■
encloses.
■
■
r 2 cos ■
■
■
■
■
■
■
bifolium. Graph it and find the area that it encloses.
■ Find the area of the region enclosed by one loop of
the curve.
19–22
20. r 3 sin 2
0 34
0 3
31. r cos 24
32. r cos 22
■
2
; 18. The curve with polar equation r 2 sin cos is called a
19. r cos 3
■
0 2
13. r 2 cos ■
■
Find the length of the polar curve.
27. r 5 cos ,
10. r 4 sin ■
(inner loop)
r sin and find the exact area of the region that lies inside
both curves.
28. r 2 ,
■
■
r 32
■
12. r 41 cos ■
■
; 26. Graph the hippopede r s1 0.8 sin 2 and the circle
; 17. Graph the curve r 2 cos 6 and find the area that it
■
■
■
■
■
■
■
■
■
■
■ Use a calculator or computer to find the length of the
loop correct to four decimal places.
33–34
33. One loop of the four-leaved rose r cos 2
34. The loop of the conchoid r 4 2 sec ■
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■
11. r sin 3
■
■
of the limaçon r 3 4 sin .
Sketch the curve and find the area that it encloses.
9. r 5 sin ■
25. Find the area inside the larger loop and outside the smaller loop
2 32
■
■
23. r 1 cos ,
0 6
■
22. r 2 3 cos ■
■ Find the area of the region that lies inside the first curve
and outside the second curve.
5. r sin 2,
8. r 2,
■
23–24
2 2
4. r 1,
■
21. r sin 5
0
3. r 2 cos ,
1
S Click here for solutions.
1– 8 ■ Find the area of the region that is bounded by the given
curve and lies in the specified sector.
2. r e ,
■
AREAS AND LENGTHS IN POLAR COORDINATES
A Click here for answers.
1. r ,
AREAS AND LENGTHS IN POLAR COORDINATES
■
■
■
■
■
■
■
■
■
■
2
■
SECTION 9.4
AREAS AND LENGTHS IN POLAR COORDINATES
9.4
ANSWERS
E Click here for exercises.
S Click here for solutions.
π
−π
2. 14 e − e
3
1. 16 π
√
3. π
+ 43
6
17.
12
4. 5π
√
3
5. 4π−3
96
1
6. 24
(π + 2)
7. 98 (π + 2)
8. 121
π5
160
9π
2
18.
9.
10.
π
8
π
19. 12
25
π
4
33π
2
11.
20. 9π
8
π
21. 20
√
22. 17
cos−1 32 − 3 5
2
23. 9 8 3 − 14 π
24. 3
√
3
√
−1 3
4
25. 34 sin
12.
√
+9 7
1.1
26.
O
1.1
1.1
24π
π
4
3
π
10
13.
(2, 0)
O
3π
2
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3, 2
16.
(4, O
19π
2
−
1
5
√
5
29. 8
1 + ln2 2 22π − 1
ln 2
√ −3π
30. 2 1 − e
31. 16
3
32. 4
33. 2.4221
34. 5.8128
27. 15
π
4
O
15.
0.6
√
arcsin 35
14.
π
−
1
10
2, 0
π
2
28.
SECTION 9.4
9.4
AREAS AND LENGTHS IN POLAR COORDINATES
SOLUTIONS
E Click here for exercises.
3 π
π 2
π 2
3
1. A = 0 21 r dθ = 0 21 θ dθ = 16 θ 0 = 16 π
11.
2θ π/2
π/2
2θ
2. A = −π/2 12 e dθ = 14 e
= 14 eπ − e−π
−π/2
π/6
π/6
(2 cos θ)2 dθ = 0 (1 + cos 2θ) dθ
√
π/6
= θ + 12 sin 2θ 0 = π6 + 43
3. A = 0
1
2
5π/6
4. A = π/6
π/6
1
2
5π/6
(1/θ)2 dθ = [−1/ (2θ)]π/6 =
π/6
sin2 2θ dθ = 14 0 (1 − cos 4θ) dθ
√
π/6
1
3
sin 4θ 0 = 4π−3
= 14 θ − 16
96
5. A = 0
π/6
sin2 3θ dθ = 3 0
π/6
= 32 θ − 16 sin 6θ 0 = π4
A=6
12
5π
π/6
0
1
2
(1 − cos 6θ) dθ
1
2
π/12
12.
π/12
cos2 3θ dθ = 12 0
(1 + cos 6θ) dθ
π/12
1
1
1
= 24 (π + 2)
= 2 θ + 6 sin 6θ 0
6. A = 2 0
3π/4
1
2
O
π/2
(3 sin θ)2 dθ = 2 π/4 94 (1 − cos 2θ) dθ
π/2
= 92 θ − 12 sin 2θ π/4 = 98 (π + 2)
7. A = π/4
3π/2
8. A = π/2
1
2
1
2
1
2
1 5 3π/2
2 2
dθ = 10
θ π/2 =
θ
=
=
121 5
π
160
9.
=
π
1
[4 (1 − cos θ)]2 dθ
0 2
π
16 0 1 − 2 cos θ + cos2 θ dθ
π
8 0 (3 − 4 cos θ + cos 2θ) dθ
4 [6θ − 8 sin θ + sin 2θ]π0 = 24π
A=2
13.
(2, 0)
A=
=
π
1
(5 sin θ)2 dθ =
0 2
π
25
θ − 12 sin 2θ 0 =
4
25
4
π
0
25
π
4
(1 − cos 2θ) dθ
π/2
(2 cos θ)2 dθ = 2
π/2
= 2 θ + 12 sin 2θ 0 = π
A=2
10.
0
1
2
π/2
0
(1 + cos 2θ) dθ
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14.
π/2
A = 2 −π/2 12 (4 − sin θ)2 dθ
π/2 = −π/2 16 − 8 sin θ + sin2 θ dθ
π/2 = −π/2 16 + sin2 θ dθ [by Theorem 5.5.7(b)]
π/2 16 + sin2 θ dθ [by Theorem 5.5.7(a)]
=2 0
π/2 16 + 12 (1 − cos 2θ) dθ
=2 0
π/2
θ − 14 sin 2θ 0 = 33π
= 2 33
2
2
π/2
A = 2 −π/2 12 (1 + sin θ)2 dθ
π/2 = −π/2 1 + 2 sin θ + sin2 2θ dθ
π/2
π/2
= [θ − 2 cos θ]−π/2 + −π/2 12 (1 − cos 2θ) dθ
π/2
= π + 12 θ − 12 sin 2θ −π/2 = π + π2 = 3π
2
■
3
4
■
SECTION 9.4
AREAS AND LENGTHS IN POLAR COORDINATES
15.
π/6
π/6
cos2 3θ dθ = 12 0 (1 + cos 6θ) dθ
π/6
π
= 12 θ + 16 sin 6θ 0 = 12
19. A = 2 0
3, 2
(4, O
1
2
2, 0
π/4
(3 sin 2θ)2 dθ =
π/4
= 92 θ − 14 sin 4θ 0 = 9π
8
20. A = 2 0
π
A = 2 0 21 (3 − cos θ)2 dθ
π
= 0 9 − 6 cos θ + cos2 θ dθ
π
= 9θ − 6 sin θ + 12 θ + 14 sin 2θ 0 =
19π
2
π/5
1
2
9
2
π/4
0
(1 − cos 4θ) dθ
π/5
sin2 5θ dθ = 14 0 (1 − cos 10θ) dθ
π/5
1
π
sin 10θ 0 = 20
= 14 θ − 10
21. A = 0
16.
1
2
22.
π/4
π/4
A = 8 0 12 sin2 4θ dθ = 2 0 (1 − cos 8θ) dθ
π/4
= 2θ − 14 sin 8θ 0 = π2
2 + 3 cos θ = 0 ⇒ cos θ = − 23 ⇒ θ = cos−1 − 23
or 2π − cos−1 − 23 . Let α = cos−1 − 23 . Then
π
A = 2 α 21 (2 + 3 cos θ)2 dθ
π
= α 4 + 12 cos θ + 9 cos2 θ dθ
π
= α 17
+ 12 cos θ + 92 cos 2θ dθ
2
π
= 17
θ + 12 sin θ + 94 sin 2θ α
2
17.
(π − α) − 12 sin α − 92 sin α cos α
√ √ = 17
π − cos−1 − 23 − 12 35 − 92 35 − 23
2
√
= 17
cos−1 32 − 3 5
2
=
By symmetry, the total area is twice the area enclosed above
the polar axis, so
π
π
A = 2 0 21 r2 dθ = 0 [2 + cos 6θ]2 dθ
π
= 0 4 + 4 cos 6θ + cos2 6θ dθ
1
π
= 4θ + 4 16 sin 6θ + 24
sin 12θ + 12 θ 0
= 4π +
π
2
=
9π
2
17
2
23.
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18.
Note that the entire curve r = 2 sin θ cos2 θ is generated by
θ ∈ [0, π]. The radius is positive on this interval, so the area
enclosed is
π
π 2
A = 0 21 r2 dθ = 0 21 2 sin θ cos2 θ dθ
π
π
= 2 0 sin2 θ cos4 θ dθ = 2 0 (sin θ cos θ)2 cos2 θ dθ
2
π
= 2 0 12 sin 2θ cos2 θ dθ
π
= 14 0 sin2 2θ (cos 2θ + 1) dθ
π
π
= 14 0 sin2 2θ cos 2θ dθ + 0 sin2 2θ dθ
π
= 14 12 θ − 14 sin 4θ 0 (the first integral vanishes) = π8
1 − cos θ = 32 ⇒ cos θ = − 12 ⇒ θ = 2π
or 4π
3
3
π 1 3 2 2
A = 2 2π/3 2 (1 − cos θ) − 2
dθ
π = 2π/3 − 54 − 2 cos θ + cos2 θ dθ
π
π
= − 54 θ − 2 sin θ 2π/3 + 12 2π/3 (1 + cos 2θ) dθ
√
π
5
π + 3 + 12 θ + 12 sin 2θ 2π/3
= − 12
√
√
√
5
= − 12
π + 3 + 16 π + 83 = 9 8 3 − 14 π
⇒
SECTION 9.4
AREAS AND LENGTHS IN POLAR COORDINATES
■
5
b
r2 + (dr/dθ)2 dθ
3π/4 (5 cos θ)2 + (−5 sin θ)2 dθ
= 0
3π/4
cos2 θ + sin2 θ dθ
=5 0
3π/4
=5 0
dθ = 15
π
4
27. L = a
24.
3 cos θ = 2 − cos θ ⇒ cos θ = 12 ⇒ θ = ± π3
π/3 A = 2 0 12 (3 cos θ)2 − (2 − cos θ)2 dθ
π/3 8 cos2 θ + 4 cos θ − 4 dθ
= 0
π/3
= 0 (4 cos 2θ + 4 cos θ) dθ
√
=3 3
= [2 sin 2θ + 4 sin θ]π/3
0
⇒
b
r2 + (dr/dθ)2 dθ
2π 2π (2θ )2 + [(ln 2) 2θ ]2 dθ = 0 2θ 1 + ln2 2 dθ
= 0
θ 2π
1 + ln2 2 22π − 1
2
2
=
1 + ln 2
=
ln 2 0
ln 2
28. L = a
π
(1 + cos θ)2 + (− sin θ)2 dθ
√ π
√ π√
2 cos2 (θ/2) dθ
= 2 2 0 1 + cos θ dθ = 2 2 0
29. L = 2 0
25.
7, 2
= [8 sin (θ/2)]π0 = 8
r 3 4 sin 3π (e−θ )2 + (−e−θ )2 dθ
√ √ 3π −θ
= 2 0 e dθ = 2 1 − e−3π
30. L = 0
3, 0
2π sin134
The curve crosses itself when 3 + 4 sin θ = 0 ⇔
sin θ = − 34 . Letting α = sin−1 34 , the desired area is
π/2
A = 2 −α 12 (3 + 4 sin θ)2 dθ
−α
− −π/2 12 (3 + 4 sin θ)2 dθ
1 θ + cos6 14 θ sin2 14 θ dθ
4
2π = 2 0 cos3 14 θ cos2 14 θ + sin2 14 θ dθ
2π = 2 0 cos3 14 θ dθ
π/2
= 8 0 cos3 u du (where u = 14 θ)
π/2
= 8 sin u − 13 sin3 u 0 = 16
3
31. L = 2 0
Now
(3 + 4 sin θ)2 dθ = 9θ − 24 cos θ + 8θ − 4 sin 2θ + C, so
√
A = 34α + 48 cos α − 16 sin α cos α = 34 sin−1 34 + 9 7 .
Note that the curve is retraced after every interval of length
4π.
1.1
26.
cos8
π 2
2 + − cos 12 θ sin 12 θ
dθ
cos2 12 θ
1 1 π
π
= 2 0 cos 2 θ dθ = 4 sin 2 θ 0 = 4
32. L = 2 0
1.1
1.1
33. From Figure 4 in Example 1,
0.6
5
9
(= α, so cos α = 23 ). So the area is
2
α
π/2 A = 2 0 21 sin2 θdθ + 2 α 12
1 − 0.8 sin2 θ dθ
α π/2
= 12 θ − 14 sin 2θ 0 + θ − 0.8 12 θ − 14 sin 2θ α
Copyright © 2013, Cengage Learning. All rights reserved.
θ = arcsin
=
1
α
2
=
1
2
−
1
4
(2 sin α cos α) + 0.6 ·
arcsin
√
5
3
π/4 r2 + (r )2 dθ
π/4 cos2 2θ + 4 sin2 2θ dθ
=2 0
L=
The points of intersection occur where
2
2
1 − 0.8 sin
θ = sin θ ⇔ 1.8 sin θ = 1 ⇔
−π/4
≈ 2 (1.211056) ≈ 2.4221
34.
π
2
− [0.6α + 0.2 (2 sin α cos α)]
−
√
1 52
2 3 3
+ 0.3π
√
−0.6 arcsin 35 − 0.4 ·
√
√
3
1
= 10
π − 10
arcsin 35 − 15 5 ≈ 0.411
√
52
3 3
4 + 2 sec θ = 0 ⇒ sec θ = −2
⇒ cos θ = − 12 ⇒ θ = 2π
, 4π
.
3
3
4π/3
L = 2π/3 (4 + 2 sec θ)2 + (2 sec θ tan θ)2 dθ ≈ 5.8128