Find the area of the region lying outside the circler= 2 but inside the circler= 4 sin

In the attached figure, you're asked to find the area of the region shaded orange. It can be found by integrating the differential of area between the two circles over the range θ = π/3 to 2π/3, or you can find it based on the geometry of the problem. (The figure shows both ways.)

I prefer to consider the geometry, rather than mess with integrals of trig functions. The area shaded blue in the figure is a circular segment with central angle 120°. The orange area is the area of the upper circle, less the area of two of those circular segments.

The area of a circular segment of central angle θ is

... A = (1/2)r²(θ - sin(θ))

In our case, we have r = 2, and θ = 2π/3. And we want to find the area of two such segments.

... (area to be subtracted from circle) = 2·(1/2)·2²·(2π/3 -sin(2π/3))

... ... = 4(2π/3 - √3/2)

... ... = 8π/3 - 2√3

This area is subtracted from the area of a circle of radius 2, which is π·2² = 4π.

Area of interest = 4π - (8π/3 -2√3) = 4π/3 + 2√3 ≈ 7.653 square units

the one that has a slope of 1/2 and is positive.

step-by-step explanation: 2 buckets 1 trip ; 4 buckets 2 trips and so on.