Calculus by Howard Anton 8th Edition

Q1. Find the area of the region inside the circle r=3sinθ and outside the cardioid r=1+sinθ.

Solution:

Area = ½∫^π/2_π/6 (3sinθ)² dθ - ½∫^π/2_π/6 (1+sinθ)² dθ

= ½ ∫^π/2_π/6 (9sin²θ) dθ - ½ ∫^π/2_π/6 (1+sin²θ) dθ

= ½∫^π/2_π/6 (9sin²θ - (1+sin²θ+2sinθ) dθ                          Simplifying 8sin²θ

= ½∫^π/2_π/6 (8sin²θ - 1 - 2sinθ) dθ                                  cos(2θ) = 1 - 2sin²(θ)   

= ½∫^π/2_π/6 (4 - 4cos(2θ) - 1 - 2sinθ) dθ                          2sin²(θ) = 1 - cos(2θ)  

= ½∫^π/2_π/6 (3 – 4cos(2θ) – 2sinθ) dθ                              8sin²(θ) = 4 - 4cos(2θ)

= ½ ∫^π/2_π/6 3dθ - 4∫^π/2_π/6 cos(2θ)dθ - 2∫^π/2_π/6 sinθdθ

= ½ (3θ - 2sin(2θ) + 2cosθ)     (π/6 to π/2)

= ½ [(3π/2 - 3π/6) – 2(sin(2π/2)-sin(2π/6) + 2(cos(π/2)-cos(π/6)]

= ½ (3π/2 - π/2) – 2(0 - √ 3/2) + 2(0 - √ 3/2)

= 2 * ½(π + √ 3/2 - √ 3/2)

= 2π/2

Area = π

Q2. State the name that describes the following polar curve most precisely: a rose, a line, a circle, a limaçon, a cardioid, a spiral, a lemniscate, or none of these
      r + 2 =2sinθ 

Also sketch and transform the given polar equation to rectangular coordinates 

Solution:

   This is CARDIOID

Polar to rectangle coordinate

r + 2 = 2sinθ

r = 2sinθ – 2

We know that

 x=rcosθ  and   y=rsinθ

Putting value of r in the above equations:

 x=(2sinθ - 2) cosθ         y=(2sinθ - 2)sinθ

 x=2sinθcosθ -2cosθ      y=2sin²θ-2sinθ

 x=sin2θ-2cosθ               y=2sinθ(sinθ-1)