Find the area of the region shared by the circle r=2 and the cardioid r=2(1-cosx)
(that x is supposed to be theta, and as you can see it's in polar mode)
To find the area of the region shared by the two curves, we need to calculate the integral of the region enclosed between the curves.
1. First, let's find the intersection points of the two curves. Equate their respective equations to find when r values are equal:
2 = 2(1 - cos θ)
Simplifying the equation, we have:
1 = 1 - cos θ
Rearranging, we get:
cos θ = 0
Therefore, the intersection points occur when θ is equal to π/2 and 3π/2.
2. Now, we need to identify the limits for our integral. Since both curves have the same r value of 2 at the intersection points, the region lies between the values π/2 and 3π/2.
3. The formula to find the area in polar coordinates is given by:
A = (1/2)∫[θ1,θ2] (r(θ))^2 dθ
In this case, r(θ) is given by 2(1 - cos θ).
4. Rewriting the integral, we have:
A = (1/2)∫[π/2,3π/2] (2(1 - cos θ))^2 dθ
5. Expanding and simplifying the equation:
A = (1/2)∫[π/2,3π/2] (4(1 - 2cos θ + cos^2 θ)) dθ
= 2∫[π/2,3π/2] (1 - 2cos θ + cos^2 θ) dθ
6. Using trigonometric identities, we can express cos^2 θ in terms of cos 2θ:
A = 2∫[π/2,3π/2] (1 - 2cos θ + (cos 2θ + 1)/2) dθ
= 2∫[π/2,3π/2] (3/2 - 2cos θ + (cos 2θ)/2) dθ
7. Distributing the integral, we get:
A = 3∫[π/2,3π/2] dθ - 4∫[π/2,3π/2] cos θ dθ + ∫[π/2,3π/2] (cos 2θ)/2 dθ
8. Integrating each term separately:
A = 3(θ)|[π/2,3π/2] - 4(sin θ)|[π/2,3π/2] + (1/4)sin 2θ|[π/2,3π/2]
= 3(3π/2 - π/2) - 4(0 - 0) + (1/4)(0 - 0)
= 6π/2
= 3π
Therefore, the area of the region shared by the circle r=2 and the cardioid r=2(1-cosx) is 3π square units.