Help me out ! Calculus question

Question

Find the  area inside the outer loop and outside the inner loop of the limacon r = 4 sin θ - 2

Answer 1

For those who wish to solve it by hand:

As stated in the video, expand r^2 = (4sinx - 2)^2 = 16sin^2(x) - 16sin(x) + 4

Therefore, A = 0.5integral(0, 2pi, 16sin^2(x) - 16sin(x) + 4, dx) - 0.5integral(pi/6, 5pi/6, 16sin^2(x) - 16sin(x) + 4, dx).

Use the fact that sin^2(x) = 0.5 - 0.5cos(2x). Resubstituting where appropriate would yield:

A = 0.5integral(0, 2pi, -8cos(2x) - 16sin(x) + 12, dx) - 0.5integral(pi/6, 5pi/6, -8cos(2x) - 16sin(x) + 12, dx).

A = (-2sin(2x) + 8cosx + 6x) from o to 2pi - (-2sin(2x) + 8cosx + 6x) from pi/6 to 5pi/6 = 35.525 (ans)

Answer 2

A graphing calculator will show r = 4sin θ − 2 as an inverted "apple" (with a dimple in the bottom for a missing stem); this graphed function has an inner and outer loop and is symmetrical to the y-axis.

The area sought will be calculated as

Area = 2[0.5∫_{(from 5π/6 to 3π/2)}(4sin θ − 2)²dθ − 0.5∫_{(from π/6 to π/2)}(4sin θ − 2)²dθ]. The area "swept" out from θ=5π/6 to θ=3π/2 is decreased by the area "swept" out from θ=π/6 to θ=π/2 and the resulting difference is then doubled to include the area to the left of the y-axis.

The necessary Algebra and Integration will give [6sin θ−4sin θcos θ+8cosθ|_{(from 5π/6 to 3π/2)}] minus

[6sin θ−4sin θcos θ+8cosθ|_{(from π/6 to π/2)}] which yields [4π+3√3] − [2π-3√3] or 2π+6√3.

Finally, double 2π+6√3 to obtain 4π+12√3 or 33.35098031 square units.