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EXAMPLE 2 Find the area of the region that lies inside the circle r= 3 sin 0 and out- side the cardioidr-1t sin0. r = 3 sin 0 SOLUTION The cardioid (see Example 10.3.7) and the circle are sketched in Fig- ure 5 and the desired region is shaded. The values of a and b in Formula 4 are deter- mined by finding the points of intersection of the two curves. They intersect when 3 sin 8 - 1+ sin 0, which gives sin e - , so 0 - w/6, 5w/6. The desired area can be found by subtracting the area inside the cardioid between # = T/6 and # = 57/6 from the area inside the circle frum 7/6 w 5n/6. Thus A = (3 sine) de - !| (1 + sin e)' de Since the region is symmetric about the vertical axis 0 - T/2, we can write 9 sin'# de –(* (1 + 2 sin e + sin*e) de - [" (8 sin'0 - 1 2 sin 0) do r=1+ sin 0 (* (3 4 cos 20 - 2 sin 6) de becnuse sin'e = t(1 - cos 28) = 30 – 2 sin 20 + 2 cos o- FIGURE 5 Set up integral that computes the area shaded in gray below. r = 3 sin 0 5т r =1+ sin 0 FIGURE 5 DO NOT EVALUATE

EXAMPLE 2 Find the area of the region that lies inside the circle r= 3 sin 0 and out- side the cardioidr-1t sin0. r = 3 sin 0 SOLUTION The cardioid (see Example 10.3.7) and the circle are sketched in Fig- ure 5 and the desired region is shaded. The values of a and b in Formula 4 are deter- mined by finding the points of intersection of the two curves. They intersect when 3 sin 8 - 1+ sin 0, which gives sin e - , so 0 - w/6, 5w/6. The desired area can be found by subtracting the area inside the cardioid between # = T/6 and # = 57/6 from the area inside the circle frum 7/6 w 5n/6. Thus A = (3 sine) de - !| (1 + sin e)' de Since the region is symmetric about the vertical axis 0 - T/2, we can write 9 sin'# de –(* (1 + 2 sin e + sin*e) de - [" (8 sin'0 - 1 2 sin 0) do r=1+ sin 0 (* (3 4 cos 20 - 2 sin 6) de becnuse sin'e = t(1 - cos 28) = 30 – 2 sin 20 + 2 cos o- FIGURE 5 Set up integral that computes the area shaded in gray below. r = 3 sin 0 5т r =1+ sin 0 FIGURE 5 DO NOT EVALUATE

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter5: Inverse, Exponential, And Logarithmic Functions
Section5.6: Exponential And Logarithmic Equations
Problem 64E
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EXAMPLE 2 Find the area of the region that lies inside the circle r= 3 sin 0 and out- side the cardioidr-1t sin0. r = 3 sin 0 SOLUTION The cardioid (see Example 10.3.7) and the circle are sketched in Fig- ure 5 and the desired region is shaded. The values of a and b in Formula 4 are deter- mined by finding the points of intersection of the two curves. They intersect when 3 sin 8 - 1+ sin 0, which gives sin e - , so 0 - w/6, 5w/6. The desired area can be found by subtracting the area inside the cardioid between # = T/6 and # = 57/6 from the area inside the circle frum 7/6 w 5n/6. Thus A = (3 sine) de - !| (1 + sin e)' de Since the region is symmetric about the vertical axis 0 - T/2, we can write 9 sin'# de –(* (1 + 2 sin e + sin*e) de - [" (8 sin'0 - 1 2 sin 0) do r=1+ sin 0 (* (3 4 cos 20 - 2 sin 6) de becnuse sin'e = t(1 - cos 28) = 30 – 2 sin 20 + 2 cos o- FIGURE 5 Set up integral that computes the area shaded in gray below. r = 3 sin 0 5т r =1+ sin 0 FIGURE 5 DO NOT EVALUATE
Transcribed Image Text:EXAMPLE 2 Find the area of the region that lies inside the circle r= 3 sin 0 and out- side the cardioidr-1t sin0. r = 3 sin 0 SOLUTION The cardioid (see Example 10.3.7) and the circle are sketched in Fig- ure 5 and the desired region is shaded. The values of a and b in Formula 4 are deter- mined by finding the points of intersection of the two curves. They intersect when 3 sin 8 - 1+ sin 0, which gives sin e - , so 0 - w/6, 5w/6. The desired area can be found by subtracting the area inside the cardioid between # = T/6 and # = 57/6 from the area inside the circle frum 7/6 w 5n/6. Thus A = (3 sine) de - !| (1 + sin e)' de Since the region is symmetric about the vertical axis 0 - T/2, we can write 9 sin'# de –(* (1 + 2 sin e + sin*e) de - [" (8 sin'0 - 1 2 sin 0) do r=1+ sin 0 (* (3 4 cos 20 - 2 sin 6) de becnuse sin'e = t(1 - cos 28) = 30 – 2 sin 20 + 2 cos o- FIGURE 5 Set up integral that computes the area shaded in gray below. r = 3 sin 0 5т r =1+ sin 0 FIGURE 5 DO NOT EVALUATE
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