Find the area of the region enclosed by one loop of the curve r=sin4Ɵ?

2 Answers

• sin4θ=0

  4θ= 0, pi, 2pi, etc

  θ= 0, pi/4, pi/2, etc

  Integrate (1/2)r^2dθ from θ= 0 to pi/4 to get one loop.

  INT (1/2)(sin4θ)^2dθ for [0, pi/4]= .196, using my TI

  🙂 R

• Calculating area using polar equation, between angles a and b:

  A = 1/2 ∫ₐᵇ r² dθ

  Now curve r = sin(4θ) has 8 loops: http://www.wolframalpha.com/input/?i=polar+plot+r+...

  One loop is located between θ = 0 and θ = π/4

  A = 1/2 ∫[θ=0..π/4] sin²4θ dθ

  A = 1/2 ∫[θ=0..π/4] 1/2 (1 - cos(8θ)) dθ

  A = 1/4 ∫[θ=0..π/4] (1 - cos(8θ)) dθ

  A = 1/4 (θ - 1/8 sin(8θ)) |[θ=0..π/4]

  A = 1/4 (π/4 - 1/8 sin(2π) - 0 + 1/8 sin(0))

  A = 1/4 (π/4)

  A = π/16