Find the area of the region that lies within thecurve r=2cosθ but is outside the curve r=1.To get full credit you need to SHOW WORK.
I don't need full credit, but you can supply any steps you think necessary.
since the region is symmetric, just double the area in the first quadrant,
The two circles intersect where
2cosθ = 1
θ = π/3
so the area is
A = 2∫[0,π/3] 1/2 (R^2-r^2) dθ
= 2∫[0,π/3] 1/2 ((2cosθ)^2-1^2) dθ = 2(π/6 + √3/4)
To find the area of the region that lies within the curve r = 2cosθ but is outside the curve r = 1, we can use the concept of polar coordinates and integration.
The region lies within the curve r = 2cosθ if the value of r for a particular angle θ is between 1 and 2cosθ. This means that the area can be divided into small sectors, and we need to find the sum of the areas of these sectors.
The area of each sector can be approximated by the formula: 0.5 * r^2 * dθ, where r is the value of r at that particular angle θ, and dθ is a small angle increment.
To find the total area, we integrate this formula for the range of θ values where r varies between 1 and 2cosθ.
∫ [0 to 2π] (0.5 * r^2 * dθ)
Since r = 2cosθ, we can substitute it in the equation:
∫ [0 to 2π] (0.5 * (2cosθ)^2 * dθ)
Simplifying the equation:
∫ [0 to 2π] (0.5 * 4cos^2θ * dθ)
= 2 ∫ [0 to 2π] (cos^2θ * dθ)
Using the identity cos^2θ = 0.5 * (1 + cos2θ), we can rewrite the equation:
= 2 ∫ [0 to 2π] (0.5 * (1 + cos2θ) * dθ)
= ∫ [0 to 2π] (1 + cos2θ) * dθ
Now we integrate this equation over the range [0 to 2π]:
∫ [0 to 2π] (1 + cos2θ) * dθ
= [θ + 0.5 * sin2θ] evaluated from 0 to 2π
= (2π + 0.5 * sin(4π)) - (0 + 0.5 * sin(0))
= 2π
Therefore, the area of the region that lies within the curve r = 2cosθ but is outside the curve r = 1 is 2π.