The circles overlap to form an area that is double that of a 120° circular segment. Thus, one can use the formula for the area of a segment (see first source link) and subtract 2 of them from the area of the circle.
.. area = (area of circle of radius 2) - (2*(area of 120° circular segment))
.. = π(2^2) - 2(2^2/2)(2π/3 - √3/2)
.. = 4π - 8π/3 + 2√3
.. = 4π/3 + 2√3 ... <== ANSWER
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This being a problem from a calculus course, one might expect to find the area by integration. In that case, the area can be considered to be the total area of a bunch of differential areas, each of which has the shape of a trapezoid. One "base" is (4sin(θ)dθ), the other "base" is 2dθ, and the height is (4sin(θ)-2). Thus, this trapezoid has area
.. d(area) = (4sin(θ)dθ + 2dθ)(4sin(θ) - 2)/2
.. = (1/2)((4sin(θ))^2 - 2^2)dθ
.. = (8sin(θ)^2 - 2)dθ
and the total area is given by the integral
.. area = ∫d(area) = ∫(8sin(θ)^2 - 2)dθ from π/6 to 5π/6
That integral gives the same result as before (see second source link).
.. area = (area of circle of radius 2) - (2*(area of 120° circular segment))
.. = π(2^2) - 2(2^2/2)(2π/3 - √3/2)
.. = 4π - 8π/3 + 2√3
.. = 4π/3 + 2√3 ... <== ANSWER
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This being a problem from a calculus course, one might expect to find the area by integration. In that case, the area can be considered to be the total area of a bunch of differential areas, each of which has the shape of a trapezoid. One "base" is (4sin(θ)dθ), the other "base" is 2dθ, and the height is (4sin(θ)-2). Thus, this trapezoid has area
.. d(area) = (4sin(θ)dθ + 2dθ)(4sin(θ) - 2)/2
.. = (1/2)((4sin(θ))^2 - 2^2)dθ
.. = (8sin(θ)^2 - 2)dθ
and the total area is given by the integral
.. area = ∫d(area) = ∫(8sin(θ)^2 - 2)dθ from π/6 to 5π/6
That integral gives the same result as before (see second source link).