Find the area of the region that lies inside the first curve and outside the second curve. r = 3 + cos(𝜃), r = 4 − cos(𝜃)

Answer 1

POLAR COORDINATES! WAHOO! CALCULUS III!

These two equations represent circles for which their centers DO NOT LIE at the origin:

1. 4 - cos(θ):
2. Circle
3. Radius 4
4. Origin shifted 1 unit to the left
5. 3 + cos(θ):
6. Circle
7. Radius 3
8. Origin shifted one unit to the right

To find the limits of integration, one must set these two equations to be equal to one another and solve for the only variable, θ:

1. 3 + cos(θ) = 4 - cos(θ)
2. 2 · cos(θ) = 1
3. cos(θ) = 1/2
4. θ = cos^-1(1/2)
5. θ = π / 3 and 5π / 3
6. or θ = ± π / 3

Now, to find the area one integrates by taking the area of the first curve and subtracting from it, the area of the second curve:

1. First curve ..... R₁ = 3 + cos(θ)
2. Second curve ..... R₂ = 4 - cos(θ)
3. Area between these two curves is given by:
4. ∫ (1/2) R₁² dθ - ∫ (1/2) R₂² dθ
5. ∫ [ (1/2) R₁² - (1/2) R₂² ] dθ
6. ∫ (1/2) [ R₁² - R₂² ] dθ
7. 1/2 · ∫ [ R₁² - R₂² ] dθ
8. 1/2 · ∫ [ ( 3 + cos(θ) )² - ( 4 - cos(θ) )² ] dθ
9. 1/2 · ∫ [ (9 + 6 cos(θ) + cos²(θ) - 16 + 8 cos(θ) - cos²(θ) ] dθ
10. 1/2 · ∫ [ -7 + 14 cos(θ) ] dθ; evaluated from - π/3 to π / 3
11. The limits of integration are∫ from - π / 3 to π / 3, and the graph is symmetrical; therefore, one may obtain the same result by integrating from 0 to π /3 and multiplying the above integral by 2 obtaining:
12. ∫ [ -7 + 14 cos(θ) ] dθ
13. ∫ -7 dθ + ∫ 14 cos(θ) dθ
14. -7 ∫ dθ + 14 ∫ cos(θ) dθ
15. -7θ + 14 sin(θ); evaluated from 0 to π/3
16. -7 (π/3) + 14 sin(π/3) + 7(0) + 14 sin(0)
17. -7π / 3 + 14 (√3 / 2)
18. (7√3 / 2) - (7π / 3)
19. ≈ 4.793972795