All right,

For part a)

We are asked to find the area of S. I assume that you have the figure of this curve plotted out. It should look like a wide squashed circle (shifted down) and smaller circle about the origin.

One way to think about the problem is to cut up the shaded area into two parts.

We know the area of a circle (constant radius) is

A=πr^{2}

We can also think of this equation as an integral of the radius

A = _{0}∫^{2π} r(θ)^{2}/2 *dθ*^{
}

If we work out the integral for a constant radius (r(θ) = R) we can prove this

A =_{ 0}∫^{2π} R^{2}/2 dθ (notice there are no θ terms in the integral, or θ^{0})

A = R^{2}/2θ |_{0}^{2π} (raise the power (0 → 1), and divide by power (/1))

A = R^{2}(0)/2 + R^{2}(2π)/2 (evaluate from 0 to 2π)

A = πR^{2 }(We are left with the equation we wanted)

Looking at our graph we see that the plots of r=3 and r = 4-2sin(θ) intersect twice. Let's find out where they intersect by setting them equal.

3 = 4-2sin(θ)

-1 = -2sin(θ)

1/2 = sin(θ)

θ = sin^{-1}(1/2) (this actually has two solutions)

θ = π/4, θ=3π/4

This solution will tell us what range for which to evaluate the area of the curves.

Looking at the graph and starting at θ = 0 and moving counter-clockwise, the shaded area is bounded by the curve r = 3 up to the point where θ = π/4.

After that, the shaded area is bounded by the curve r = 4-2sin(θ) from θ = π/4 to θ = 3π/4

After that, the shaded area is again bounded by the curve r = 3 from θ = 3π/4 all the way back around to θ = 2π.

So we can find the area of the shaded region by adding up the following areas

A_{1} = _{0}∫^{π/4} 3^{2}/2 *d*θ

A_{2} = _{π/4}∫^{3π/4} [4-2sin(θ)]^{2}/2 *d*θ

A_{3} = _{3π/4}∫^{2π} 3^{2}/2 dθ

Let's solve them one at a time

A_{1} = 9θ/2 |_{0}^{π/4}

A_{1} = 0 + 9/2*π/4 = **9π/8**

A_{2 = } _{ π/4}∫^{3π/4} (4-2sin(θ)) (4-2sin(θ))/2 *d*θ

A_{2} = _{π/4}∫^{3π/4} (16-16sin(θ)+4sin^{2}(θ))/2 dθ

A_{2} =_{ π/4}∫^{3π/4} 8-8sin(θ)+2sin^{2}(θ) dθ

A_{2} =_{ π/4}∫^{3π/4} 8 -_{ π/4}∫^{3π/4} 8sin(θ) + _{π/4}∫^{3π/4} 2sin^{2}(θ)

A_{2} = 8θ + 8cos(θ) + 2/2 (θ-sin(θ)cos(θ)) |_{π/4}^{3π/4 }(evaluate this at π/4 to 3π/4)

A_{2 =} 8π/4 + 8cos(π/4) + π/4-sin(π/4)cos(π/4) + 24π/4 + 8cos(3π/4) + 3π/4-sin(3π/4)cos(3π/4)

A_{2 = }8π/4 + 8√2/2 + π/4-√2/2*√2/2 + 24π/4 - 8√2/2 + 3π/4+√2/2*√2/2

A_{2 =} (8+1+24+3)π/4 = **6π**

A_{3 = 3π/4}∫^{2π }3^{2}/2 *d*θ

A_{3 = }9θ/2 |_{3π/4}^{2π}

A_{3} = 9*3π/(2*4) + 9*2π/2

A_{3 = }**27π/8 + 9π**

Add them all together and we get

A_{1} + A_{2} + A_{3} = 9π/8 + 6π + 27π/8 + 9π = 9π/2+15π =
**39π/2**

This is not correct. I made an error right at the beginning. The two curves intersect at π/6 and 5π/6

## Comments

For part b, you say set (4-2sin(θ))cos(θ) = -1 where θ = t

^{2}and t is in the interval [1,2]. However, I don't understand exactly how you are solving this either graphically or on a calculator.Comment