Area of Polar Curve: Find the Area Enclosed by r=2+sin(4θ)

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Homework Help Overview

The discussion revolves around finding the area enclosed by the polar curve defined by the equation r=2+sin(4θ). Participants are exploring the appropriate limits of integration for calculating the area.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to determine the limits of integration by setting sin(4θ)=0, leading to the values 0 and π/4, but they express confusion over the correctness of these limits. Other participants suggest plotting the curve to gain insights into the problem and question the relationship between the period of r(θ) and the limits.

Discussion Status

Participants are actively questioning the limits of integration and the period of the function. There is acknowledgment of a potential misunderstanding regarding the period of the polar curve, with some participants agreeing that the initial limits may not be appropriate.

Contextual Notes

There is a mention of the curve being positive for all θ, which raises questions about how the limits relate to the overall area calculation. The correct answer is noted to be 4.5π, but the reasoning behind the limits remains under discussion.

steel1
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Homework Statement


Find the area enclosed by the graph r=2+sin(4θ)


Homework Equations


Area = .5∫r^2



The Attempt at a Solution


Area = .5∫(2+sin(4θ))^2
=.5(4.5θ-1/16sin(8θ)-cos(4θ))

I can do the integration and all, but I am having trouble finding the limits of integration

I set sin(4θ)=0 and solve, and get 0 and pi/4. but when i use these limits, i get the wrong answer.

correct answer is 4.5pi
 
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Have you tried plotting the curve (either using a calculator or computer, or by hand)?
That should give you a clue (hint: the correct answer is related to the period of r(theta)).
 
CompuChip said:
Hhint: the correct answer is related to the period of r(theta)).
It is? i don't see that. Since r > 0 for all theta, the curve cannot join up until it has gone all the way around 2π.
 
Exactly. So why the pi / 4?
 
CompuChip said:
Exactly. So why the pi / 4?

I agree the pi/4 is wrong, but the period of r would be pi/2, which is also wrong.
 
Ah, you are right, the problem was in my wording. Thanks!
 

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