Calculate the area outside r = sinθ and inside r = 2sinθ?

Answer 1

The area element in polar coordinates is r dr dθ.
So we have A = ∫(0 to 2π) ∫(sin θ to 2sin θ) r dr dθ
= ∫(0 to 2π) ((2 sin θ)^2/2 - (sin θ)^2/2)dθ
= ∫(0 to 2π) (3/2) sin^2 θ dθ
= ∫(0 to 2π) (3/4) (1 - cos 2θ) dθ
= 3/4 (2π) - 3/4 [(sin 2θ)/2][0 to 2π]
= 3π/2 - 3/8(0 - 0)
= 3π/2.

Answer 2

T = theta
First find intersection sinT = 2 sinT
sinT = 0 ====> T = 0 and pi

Area in polar = 1/2 * int r^2 dT

A = 1/2 * int (0 to pi) (2sinT)^2 - (sinT)^2 dT
A = 1/2 * int (0 to pi) 3 (sinT)^2 dT
to integrate sin^2(T) = (1-cos2T) / 2 = 1/2 - cos2T / 2
A = 3/4 *int ( 1 - cos2T ) dT
integrate
A = 3/4 [ T - (sin2T) / 2 ] eval from 0 to pi

A =3/4 ( [pi - sin 2pi ] - [0 - (sin 0) / 2] )
A = 3/4 (pi)

Area is 3/4 (pi)

:)

SHORT CUT:
Outside the circle with radius 1/2
inside the circle radius 1
A = pi*(1)^2
A = pi*(1/2)^2

pi - 1/4*pi = 3/4 * pi

:)