Finding Limits for Polar Coordinate Area Integration

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Discussion Overview

The discussion revolves around finding limits for integration in polar coordinates, specifically in the context of calculating areas bounded by polar curves and lines. Participants explore the selection of appropriate limits for integration and the implications of the polar equations involved.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant seeks guidance on determining the correct limits for integration when calculating the area between the curve r=2cos(2t) and the half-line t=pi/6.
  • Another participant suggests finding the values of theta where r becomes zero to establish the limits of integration, proposing to use the first quadrant value as the upper bound.
  • A question is raised about the necessity of finding the value of theta at r=0, indicating a need for clarity on its significance.
  • A further inquiry emphasizes the importance of sketching the graph to better understand the area being calculated.
  • One participant shares their working on a similar problem involving the curve r=3cos(2θ) and requests verification of their method and calculations.
  • Another participant questions the consistency of the coefficients used in the equations presented in different posts, highlighting potential discrepancies in the problem setup.
  • The original poster clarifies that their initial equation was hypothetical and later aligned with a textbook problem, expressing appreciation for the assistance received.

Areas of Agreement / Disagreement

Participants express varying levels of understanding regarding the selection of limits for integration, with some proposing methods while others seek clarification. There is no consensus on the best approach, and multiple viewpoints are presented without resolution.

Contextual Notes

Participants reference specific polar equations and limits, but there are indications of potential arithmetic errors and inconsistencies in the equations discussed. The discussion remains focused on the methodology rather than definitive solutions.

Who May Find This Useful

Students and individuals interested in polar coordinate integration, particularly those seeking to understand the process of determining limits for area calculations in polar coordinates.

daster
I need help with finding areas. I'm having trouble picking the correct limits for my integration.

Say for example we had r=2cos2t and a half-line t=pi/6, and I want to find the small area bounded between them.

[tex]\frac{1}{2}\int (2\cos (2\theta))^{2}\,d\theta[/tex]

I can do the integration, but what do I choose as its limits?

I'm not particularly interested in the answer to this question; I'm looking for a good explanation or maybe a couple of pointers.
 
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At what value(s) of theta does [itex]r[/itex] become zero ? Set the r(theta) equation to zero and solve for theta. Use the 1st quadrant value of theta obtained as the upper bound with [itex]\frac{\pi}{6}[/itex] as the lower.
 
Why do we want the value of theta at r=0?
 
daster said:
Why do we want the value of theta at r=0?

Have you even sketched the graph yet ? It will become clear as day if you do.
 
Can you please check my working?

(a) Sketch the curve with polar equation

[tex]r=3\cos 2\theta, \, -\frac{\pi}{4}\leq\theta<\frac{\pi}{4}[/tex].

The curve looks like 1 rose petal.

(b) Find the area of the smaller finite region enclosed between the curve and the half-line [itex]\theta=\frac{\pi}{6}[/itex].

[tex]Area = \frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} (3\cos 2\theta)^2\,d\theta = \frac{9}{4}\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} (1+\cos 4\theta)\,d\theta = \frac{9}{4}\left[\theta+\frac{1}{4}\sin 4\theta\right]_{\frac{\pi}{6}}^{\frac{\pi}{4}} = \frac{3\pi}{16}- \frac{9\sqrt{3}}{32}[/tex]

A couple of arithmetic slips probably made it through, since I did all of this using my keyboard & notepad.exe. :-p So I'm just wondering if my method was correct.
 
Last edited by a moderator:
The coefficient is 3 ? Not 2 ? If that's the case then your working in the last post is correct. But why did you give a different equation in your first post ? :confused:
 
In my first post I made up an equation and a line, and when I went to do a question from my textbook, it happened to be almost like the one I made up. :-p

Anyway, thanks for your help. I appreciate it. :smile:
 
Sure. :smile:
 

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