Find the area of the region enclosed by one loop of the curve r=sin4Ɵ?
I am trying to find the area of the region enclosed by one loop of the curve given by the polar equation ( r = \sin(4\theta) ). I understand that this is a problem involving area and arc length in polar coordinates. However, I am confused about the role of the coefficient 4 in the equation. I know that the angles where ( \sin ) equals 0 are at ( 0 ) and ( \pi ). Should I multiply the 4 by ( \pi ) and 0 to determine the limits of integration? Any guidance on how to approach this problem would be greatly appreciated.
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
Feb 08, 2025
Calculating area using polar equation, between angles a and b:
A = 1/2 â«ₐáµ 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 = Ï/4
A = 1/2 â«[θ=0..Ï/4] sin²4θ dθ
A = 1/2 â«[θ=0..Ï/4] 1/2 (1 - cos(8θ)) dθ
A = 1/4 â«[θ=0..Ï/4] (1 - cos(8θ)) dθ
A = 1/4 (θ - 1/8 sin(8θ)) |[θ=0..Ï/4]
A = 1/4 (Ï/4 - 1/8 sin(2Ï) - 0 + 1/8 sin(0))
A = 1/4 (Ï/4)
A = Ï/16
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