Polar area?

Question

Find the area of the region inside the graph of the polar curve
r = 2 and outside the graph of the polar curve r = 2(1-sinθ).

r = 2 is a circle with radius of 2 and is centered at the origin
r = 2(1-sinθ) is a cardioid (heart-shaped) that lies mostly below the x-axis.
The two graphs intersect at θ = 0 & pi

Answer 1

Polar Area = 1/2 int(A to B)r^2 dT ===> T= theta
A = 1/2 int(0 upto pi) [(2)^2 - (2-2sinT)^2] dT
A = 1/2 int [ 4 -( 4 - 8sinT + 4sin^2(T) ] dT
= int [ 2 sinT + 2 sin^2(T) ] dT ====> trig id sin^2(T) = (1-cos(2T) ) /2

= int [ 2 sinT + 1 - cos(2T) ] dT
= -2cosT + T - sin(2T) / 2 [evaluate from 0 to pi]
= -2cospi + pi- sin(2pi) / 2 - [-2cos0 + 0 - sin(2*0) / 2 ]
= 2 + pi - 0 + 2 - 0 + 0
= pi + 4

AREA enlcosed = pi + 4

=)

Answer 2

A. Polar bears= 3:20 Penguins= 12:20 = 3:5 (on simplification) Reindeer= 5:20 = a million:4 B. Polar bears = 3/20 * 360 = fifty 4 deg Penguins = 12/20 * 360 = 216 deg Reindeer = 5/20 * 360 = ninety deg C. Use the angles won from (B) to create a circle graph.

Answer 3

The circle has a radius that is greater or equal to the cardiod from θ = 0 to π. So integrate over that interval.

Area = ∫[½(R² - r²)]dθ = ½∫{2² - [2(1 - sinθ)]²} dθ

= 2∫{1 - [1 - 2sinθ + sin²θ]} dθ

= 2∫{1 - 1 + 2sinθ - sin²θ} dθ

= 2∫{2sinθ - sin²θ} dθ = 2∫{2sinθ - (1 - cos2θ)/2} dθ

= 2∫{2sinθ - ½ + ½cos2θ} dθ

= 2[-2cosθ - θ/2 + ¼sin2θ] | [Evaluated from 0 to π]

= 2[2 - π/2 + 0] - 2[-2 - 0 + 0] = 4 - π + 4 = 8 - π