area between curves question

Question

Find the area that is inside r=3+3sin θ and outside r=2

Answer 1

Let find intersaction of this lines: 3 + 3sinθ = 2, from here sinθ = - 1/3; θ = - sin^-1(1/3), or θ = π + sin^-1(1/3).

So area A = 1/2∫[(3 + 3sinθ)² - 2²]dθ from θ = - sin^-1(1/3) to θ = π + sin^-1(1/3);

A = 1/2∫(9 + 18sinθ+ 9sin²θ - 4)dθ = 1/2∫(5 +18sinθ + 9/2 - 9/2 cos2θ)dθ = 1/2(19/2θ - 18cosθ - 9/4sin2θ) =

= 2·1/2[(19/2(π/2) - 18cos(π/2) - 9/4sin(π)) - (- 19/2sin^-1(1/3) - 18cos(-sin^-1(1/3)) - 9/4sin(-2sin^-1(1/3))] =

19π/4 + 19/2sin^-1(1/3) + 12sqrt(2) - 9/4·2·1/3·2sqrt(2)/3 = 19π/4 + 19/2sin^-1(1/3) + 12sqrt(2) - sqrt(2);

A = 19π/4 + 19/2sin^-1(1/3) + 11sqrt(2) ≈ 33.71;

We use the symmetry about axis y.

Answer 2

The area of the unit cardiod is 3π/2 (see https://www.whitman.edu/mathematics/calculus_online/section10.03.html)

Thus, the area inside 3 + 3 sin θ is 3^2 * 3π/2 = 27π/2

The area inside a circle with radius 2 is 4π

Thus, the area between is 27π/2 - 4π = 19π/2