Section 10.4 Calculus in Polar Coordinates 10. Sketch the curve and ﬁnd the area that it encloses.

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Section 10.4 Calculus in Polar Coordinates

10. Sketch the curve and find the area that it encloses. r = 2 + 2 cos θ.

Solution:

SECTION 10.4 CALCULUS IN POLAR COORDINATES ¤ 989 7.  = 4 + 3 sin , −^2 ≤  ≤ ^2.

 =

 2

−2 1

2(4 + 3 sin )² = ¹₂

 2

−2

(16 + 24 sin  + 9 sin²) 

= ¹₂

 2

−2

(16 + 9 sin²)  [by Theorem 5.5.7]

= ¹₂· 2

 2 0

16 + 9 ·¹2(1 − cos 2)

 [by Theorem 5.5.7]

=

 2 0

41

2 −⁹2cos 2

 =41

2 −⁹4sin 22 0 =41

4 − 0

− (0 − 0) = ^414

8.  =√

ln , 1 ≤  ≤ 2.

 =

 2

1 1 2

√ln 2

 =

 2

1 1

2ln   =

1

2 ln 2

1 −

 2

1 1 2

  = ln ,  = ¹₂

 = (1) ,  =¹₂



= [ ln(2) − 0] −

1 22

1 =  ln(2) −  + ¹2

9. The area is bounded by  = 4 cos  for  = 0 to  = .

 =

  0

1

2² =

  0

1

2(4 cos )² =

  0

8 cos² 

= 8

  0

1

2(1 + cos 2)  = 4

 +¹₂sin 2 0 = 4

Also, note that this is a circle with radius 2, so its area is (2)²= 4.

10.  = 2

0 1

2² =

 2

0 1

2(2 + 2 cos )²

=

 2

0 1

2(4 + 8 cos  + 4 cos²) 

=

 2

0 1 2

4 + 8 cos  + 4 ·¹2(1 + cos 2)



=

 2

0

(3 + 4 cos  + cos 2)  =

3 + 4 sin  +¹₂sin 22

0 = 6

11.  = 2

0 1

2² =

 2

0 1

2(3 − 2 sin )²

= ¹₂

 2

0 (9 − 12 sin  + 4 sin²) 

= ¹₂

 2

0

9 − 12 sin  + 4 ·¹2(1 − cos 2)



= ¹₂

 2

0 (11 − 12 sin  − 2 cos 2)  = ¹2

11 + 12 cos  − sin 22

0

= ¹₂[(22 + 12) − 12] = 11

28. Find the area of the region that lies inside the first curve and outside the second curve. r = 3 sin θ, r = 2 − sin θ.

Solution:

SECTION 10.4 CALCULUS IN POLAR COORDINATES ¤ 993

27. 3 cos  = 1 + cos  ⇔ cos  = ¹2 ⇒  = ^3 or −^₃.

 = 23 0

1

2[(3 cos )²− (1 + cos )²] 

=3

0 (8 cos² − 2 cos  − 1)  =3

0 [4(1 + cos 2) − 2 cos  − 1] 

=3

0 (3 + 4 cos 2 − 2 cos )  =

3 + 2 sin 2 − 2 sin 3 0

=  +√ 3 −√

3 = 

28. 3 sin  = 2 − sin  ⇒ 4 sin  = 2 ⇒ sin  = ¹2 ⇒  = ^6 or ^5₆.

 = 22

6 1

2[(3 sin )²− (2 − sin )²] 

=2

6 (9 sin² − 4 + 4 sin  − sin²] 

=2

6 (8 sin² + 4 sin  − 4) 

= 42

6

2 ·¹2(1 − cos 2) + sin  − 1



= 42

6 (sin  − cos 2)  = 4

−cos  −¹2sin 22

6

= 4

(0 − 0) −

−^√2³−^√4³

= 4

3√ 3 4

= 3√ 3

29. 3 sin  = 3 cos  ⇒ 3 sin 

3 cos  = 1 ⇒ tan  = 1 ⇒  = ^4 ⇒

 = 2

 4 0

1

2(3 sin )² =

 4 0

9 sin²  =

 4

0 9 ·¹2(1 − cos 2) 

=

 4 0

9

2 −⁹2cos 2

 =

9

2 −⁹4sin 24 0 =9

8 −⁹4

− (0 − 0)

= ^9₈ −⁹4

30.  = 42 0

1

2(1 − cos )² = 22

0 (1 − 2 cos  + cos²) 

= 22 0

1 − 2 cos  +¹2(1 + cos 2)



= 22 0

3

2− 2 cos  +¹2cos 2

 =2

0 (3 − 4 cos  + cos 2) 

=

3 − 4 sin  + ¹2sin 22

0 = ^3₂ − 4

31. sin 2 = cos 2 ⇒ sin 2

cos 2 = 1 ⇒ tan 2 = 1 ⇒ 2 = ^4 ⇒

 =^₈ ⇒

 = 8 · 2

 8 0

1

2sin²2  = 8

 8 0

1

2(1 − cos 4) 

= 4

 − ¹4sin 48 0 = 4_

8 −¹4· 1

= ^₂ − 1

45. Find the area of the shaded region.

Solution:

SECTION 10.4 CALCULUS IN POLAR COORDINATES ¤ 997

45.1 + cos  = 3 cos  ⇒ 1 = 2 cos  ⇒ cos  =¹2 ⇒  =^3. The area swept out by  = 1 + cos , 3 ≤  ≤ , contains the shaded region plus the portion of the circle  = 3 cos , 3 ≤  ≤ 2. Thus, the area of the shaded region is given by

 =



3 1

2(1 + cos )² −

 2

3 1

2(3 cos )² =¹₂



3

(1 + 2 cos  + cos²)  −⁹2

2

3

cos² 

=¹₂

 

3

1 + 2 cos  +¹₂(1 + cos 2)

 −⁹2

2

3 1

2(1 + cos 2) 

=¹₂

 

3

3

2+ 2 cos  +¹₂cos 2

 −⁹4

2

3

(1 + cos 2) 

=¹₂3

2 + 2 sin  +¹₄sin 2

3−⁹4

 +¹₂sin 22

3

=¹₂

3

2 −

 2+√

3 +^√₈³

−⁹4

 2−

 3+^√₄³

=¹₂

 −⁹^√8³



−⁹4

 6−^√4³



=^₂−^38 =^₈

46.The pole is reached when  = 1 − 2 sin  = 0 ⇒ sin  = ¹2 ⇒  =^6,^5₆, or^13₆ . The curve’s inner loop is traced from  = 6 to  = 56 (corresponding to negative values), while the outer loop is traced from  = 56 to  = 136.

From the figure, we see that the area of the outer loop minus the area of the inner loop gives the area of the shaded region.

Thus,

 =

136 56

1

2(1 − 2 sin )² −

56

6 1

2(1 − 2 sin )²

=

136 56

1

2(1 − 4 sin  + 4 sin²)  −

56

6 1

2(1 − 4 sin  + 4 sin²) 

=

136 56

1

2− 2 sin  + 2 ·¹2(1 − cos 2)

 −

 56

6

1

2− 2 sin  + 2 ·¹2(1 − cos 2)



=

1

2 + 2 cos  +  −¹2sin 2136 56 −

1

2 + 2 cos  +  −¹2sin 256

6

=

13

12 +√

3 +^13₆ −^√4³



−

5

12−√

3 +^5₆ +^√₄³

−

5

12−√

3 +^5₆ +^√₄³ +

 12+√

3 +^₆−^√4³



=

13

4 +³^√₄³

− 2

5

4 −³^√4³

 +

 4+³^√₄³

=  + 3√ 3 47.

From the first graph, we see that the pole is one point of intersection. By zooming in or using the cursor, we find the values of the intersection points to be  ≈ 088786 ≈ 089 and  −  ≈ 225. (The first of these values may be more easily estimated by plotting  = 1 + sin  and  = 2 in rectangular coordinates; see the second graph.) By symmetry, the total

1

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53. Find the exact length of the portion of the curve shown in blue.

45.

46.

囝4衍7. The p仰仰O叫int臼sofin叫nt帥ter即ction叫∞noft伽h悅1唸ec臼a吋Oωi泊dr = 1

+

sin口in叫lÐand the spiralloop r = 2Ð，一 τ/2 廷。逗 τ/2 ，can't be found exactly. Use a graph to find the approximate values of Ð at which the curves intersect. Then use these values to estimate the area that lies inside both curves.

48. When recording live performances, sound engineers often use a microphone with a cardioid pickup pattern because it suppresses noise from the audience. Suppose the microphone is placed 4 m from the front of the stage (as in the figure) and the boundary of the optimal pickup region is given by the cardioid r = 8

+

8 sin Ð, where r is measured in meters and the microphone is at the pole. The musicians want to know the area they wiU have on stage within the optimal pickup range of the microphone. Answer their question.

stage

audience

49-52 Find the exact length of the polar curve.

49. r = ²cose，。三 Ð ::::三 τ 50. r = e^O/立， 0 ::::;; Ð 三 τ/立

51. r = Ð元。三三 Ð ::::;; 27T 52. r = 2(1 + cosθ)

SECTION 10.4 Calculus in Polar Coordinates 701

53-54 Find the exact length of the portion of the curve shown in blue.

53.

r = 3

+

3 sin ()

54.

回55-56 Find the exact length of the curve. U se a graph to

dete位rm

55. r = cos⁴(Ð/4) 56. r = cos²(Ð/2)

57-58 Set up, but do not evaluate, an integral to find the length of the portion of the curve shown in blue.

57.

58.

r = cos( 6/5)

sin 6

r= 6

囝

59

-62

Use

a 叫culator

or computer to find the length of the curve co叮ectto four decimal places. If necessary, graph the curve to determine the parameter interval.

59. One loop of the curve r = cos 2Ð 60. r = tanÐ ，付/6 乏。石 π/3

61. r = sin(6 sinÐ) 62. r = sin(Ð /4) Solution:

SECTION 10.4 CALCULUS IN POLAR COORDINATES ¤ 999 51.  = 



²+ ()² =

 2

0



(²)²+ (2)² =

 2

0

⁴+ 4²

=

 2

0



²(²+ 4)  =

 2

0



²+ 4 

Now let  = ²+ 4, so that  = 2  

  = ¹₂ and

 2

0



²+ 4  =

 4²+4 4

1 2

√  = ¹₂· ²3

^324(²+1)

4 = ¹₃[4^32(²+ 1)^32− 4^32] = ⁸₃[(²+ 1)^32− 1]

52.  =

 



²+ ()² =

 2

0

[2(1 + cos )]²+ (−2 sin )² =

 2

0

4 + 8 cos  + 4 cos² + 4 sin² 

=

 2

0

√8 + 8 cos   =√ 8

 2

0

√1 + cos   =√ 8

 2

0



2 ·¹2(1 + cos ) 

=√ 8

 2

0

 2 cos² 

2 =√ 8√

2

 2

0



cos 2



  = 4 · 2

  0

cos

2 [by symmetry]

= 8

 2 sin

2

 0

= 8(2) = 16

53. The blue section of the curve  = 3 + 3 sin  is traced from  = −2 to  = .

²+ ()²= (3 + 3 sin )²+ (3 cos )²= 9 + 18 sin  + 9 sin² + 9 cos² = 18 + 18 sin 

 =

 

−2

√18 + 18 sin   =√ 18

 

−2

√1 + sin   =√ 18

 

−2

(1 + sin )(1 − sin ) 1 − sin  

=√ 18

 

−2



1 − sin²

1 − sin   =√ 18

 

−2

 cos²

1 − sin  =√ 18

 

−2

|cos |

√1 − sin 

=√ 18

 2

−2

cos 

√1 − sin  +

 

2

− cos 

√1 − sin 



=s √ 18

 1

−1

√ 1

1 −  −

 0 1

√ 1

1 − 



=√ 18

−2(1 − )^121

−1−

−2(1 − )^120 1



=√ 18

2√ 2 + 2

= 12 + 6√ 2

54. The blue section of the curve  =  + 2 is traced from  = 0 to  = 3.

²+ ()²= ( + 2)²+ (1)² = ( + 2)²+ 1

 =

 3

0

( + 2)²+ 1 

=

 tan⁻¹(3+2) tan⁻¹2

tan² + 1 sec² 

 + 2 = tan 

 = sec² 



=

 tan⁻¹(3+2) tan⁻¹2

√sec² sec²  =

 tan⁻¹(3+2)

tan⁻¹2 |sec | sec² 

=

 tan⁻¹(3+2) tan−12

sec³  = ¹₂

sec  tan  + ln |sec  + tan |tan⁻¹(3+2)

tan−12 [by Example 7.2.8]

= ¹₂

1 + (3 + 2)²(3 + 2) + ln

1 + (3 + 2)²+ 3 + 2

−¹2

√5 (2) + ln√

5 + 2

[ 6412]

68. Find the s1ope of the tangent 1ine to the given polar curve at the point specified by the value of θ.

r = 1 + 2 cos θ, θ =π 3 Solution:

1002 ¤ CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES

67.  = cos 2 ⇒  =  cos  = cos 2 cos ,  =  sin  = cos 2 sin  ⇒



= 

 = cos 2 cos  + sin  (−2 sin 2) cos 2 (− sin ) + cos  (−2 sin 2) When  = 

4, 

= 0√

22 +√

22 (−2) 0

−√ 22

+√

22

(−2) = −

√2

−√ 2 = 1.

68.  = 1 + 2 cos  ⇒  =  cos  = (1 + 2 cos ) cos ,  =  sin  = (1 + 2 cos ) sin  ⇒



= 

 = (1 + 2 cos ) cos  + sin  (−2 sin ) (1 + 2 cos )(− sin ) + cos  (−2 sin ) When  = 

3, 

= 2₁

2

+√

32

−√ 3 2

−√ 32

+¹₂

−√ 3 ·2

2 = 2 − 3

−2√ 3 −√

3 = −1

−3√ 3 =

√3 9 .

69.  = sin  ⇒  =  cos  = sin  cos ,  =  sin  = sin² ⇒ ^ = 2 sin  cos  = sin 2 = 0 ⇒ 2 = 0or  ⇒  = 0 or^2 ⇒ horizontal tangent at (0 0), and

1 ^₂ .



 = − sin² + cos² = cos 2 = 0 ⇒ 2 = ^2 or ^3₂ ⇒  = ^4 or ^3₄ ⇒ vertical tangent at

√1 2^₄ and

√1 2^3₄ 

.

70.  = 1 − sin  ⇒  =  cos  = cos  (1 − sin ),  =  sin  = sin  (1 − sin ) ⇒



 = sin  (− cos ) + (1 − sin ) cos  = cos  (1 − 2 sin ) = 0 ⇒ cos  = 0 or sin  = ¹2 ⇒

 = ^₆,^₂, ^5₆, or ^3₂ ⇒ horizontal tangent at1 2^₆,1

2^5₆ , and 2^3₂.



 = cos  (− cos ) + (1 − sin )(− sin ) = − cos² − sin  + sin² = 2 sin² − sin  − 1

= (2 sin  + 1)(sin  − 1) = 0 ⇒

sin  = −¹2 or 1 ⇒  = ^76 ,^11₆ , or^₂ ⇒ vertical tangent at₃

2^7₆ 

₃

2^11₆  , and

0^₂ . Note that the tangent is vertical, not horizontal, when  = ^₂, since

lim

→(2)⁻



 = lim

→(2)⁻

cos  (1 − 2 sin )

(2 sin  + 1)(sin  − 1) = ∞ and lim

→(2)⁺



 = −∞.

71.  = 1 + cos  ⇒  =  cos  = cos  (1 + cos ),  =  sin  = sin  (1 + cos ) ⇒



 = (1 + cos ) cos  − sin² = 2 cos² + cos  − 1 = (2 cos  − 1)(cos  + 1) = 0 ⇒ cos  = ¹2 or −1 ⇒

 = ^₃, , or ^5₃ ⇒ horizontal tangent at₃

2^₃, (0 ), and₃

2^5₃ .



 = −(1 + cos ) sin  − cos  sin  = − sin  (1 + 2 cos ) = 0 ⇒ sin  = 0 or cos  = −¹2 ⇒

 = 0, , ^2₃ , or ^4₃ ⇒ vertical tangent at (2 0),₁

2^2₃  , and₁

2^4₃ . Note that the tangent is horizontal, not vertical when  = , since lim

→



 = 0.

72.  = ^ ⇒  =  cos  = ^cos ,  =  sin  = ^sin  ⇒



 = ^sin  + ^cos  = ^(sin  + cos ) = 0 ⇒ sin  = − cos  ⇒ tan  = −1 ⇒

 = −¹4 +  [ any integer] ⇒ horizontal tangents at

^(^−14) 

 −¹4

.

[continued]

2