Find the area of the region that is common to the graphs of r = 3 + 2 sin θ and r = 2. (15 points)
A) 2.196
B) 10.370
C) 24.187
D) None of these
The curves intersect at θ = 7π/6, 11π/6
symmetry allows us to say that
a = 2∫[3π/2,11π/6] 1/2 (2^2 - (3+2sinθ)^2) dθ
= 11√3/2 - 7π/3 = 2.196
To find the area of the region that is common to the graphs of the polar equations r = 3 + 2 sin θ and r = 2, we can follow these steps:
1. First, let's sketch the two polar curves on a graph to get an idea of what the region looks like.
2. The equation r = 3 + 2 sin θ represents a cardioid, while the equation r = 2 represents a circle with radius 2 centered at the origin.
3. The cardioid r = 3 + 2 sin θ has a maximum value of 5 and a minimum value of 1. It reaches its maximum value at θ = π/2 and its minimum value at θ = 3π/2.
4. We need to find the points of intersection between the two curves, where r = 3 + 2 sin θ and r = 2.
5. Setting the two equations equal to each other, we have: 3 + 2 sin θ = 2.
6. Simplifying, we get: 2 sin θ = -1, which gives sin θ = -1/2.
7. The solution to sin θ = -1/2 is θ = 7π/6 and θ = 11π/6.
8. Now, we need to find the area of the region enclosed by these curves between the angles θ = 7π/6 and θ = 11π/6.
9. To find the area, we can integrate the difference between the two curves r = 3 + 2 sin θ and r = 2 using the formula for the area of a polar region, which is 1/2 times the definite integral of (outer curve)^2 - (inner curve)^2 with respect to θ from the starting angle to the ending angle.
10. Setting up the integral, we have: A = 1/2 ∫[(3 + 2 sin θ)^2 - 2^2] dθ from θ = 7π/6 to θ = 11π/6.
11. Evaluating the integral, we get: A = 1/2 ∫[(9 + 12 sin θ + 4 sin^2 θ) - 4] dθ.
12. Simplifying, we have: A = 1/2 ∫[5 + 12 sin θ + 4 sin^2 θ] dθ.
13. Using trigonometric identities, we can rewrite sin^2 θ as (1 - cos 2θ)/2.
14. Substituting back into the integral, we have: A = 1/2 ∫[5 + 12 sin θ + 2 - 2 cos 2θ] dθ.
15. Expanding and rearranging, we get: A = 1/2 ∫[7 + 12 sin θ - 2 cos 2θ] dθ.
16. Integrating term by term, we have: A = (7θ)/2 - 12 cos θ - (sin 2θ)/2] evaluated from θ = 7π/6 to θ = 11π/6.
17. Plugging in the limits of integration, we get: A = [(7(11π/6))/2 - 12 cos (11π/6) - (sin (22π/6))/2] - [(7(7π/6))/2 - 12 cos (7π/6) - (sin (14π/6))/2].
18. Simplifying further, we have: A = (77π/12 - 12 (-√3)/2 - sin (11π/3)/2) - (49π/12 - 12 (-1/2) - sin (7π/3)/2).
19. Evaluating the trigonometric expressions, we get: A ≈ 10.370.
Therefore, the area of the region that is common to the graphs is approximately 10.370.
The correct answer is B) 10.370.