What is the Area Between Two Polar Curves?

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Homework Help Overview

The problem involves finding the area between two polar curves: \( r = 9\sin\theta \) and \( r = 2 \). The original poster attempts to set up the integral for the area but expresses confusion regarding the limits of integration and the correctness of their calculations.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the limits of integration, with some questioning whether to integrate from the intersection point to \( \pi/2 \). There is also a suggestion to sketch the curves to better understand the area in question.

Discussion Status

Participants are exploring different approaches to determine the correct limits of integration and the setup for the area calculation. Some guidance is offered regarding the use of the arcsin function to find intersection points, and there is acknowledgment of the symmetry of the shaded region.

Contextual Notes

There is mention of potential rounding errors when substituting values for theta, and participants are considering the implications of symmetry in their calculations. The discussion reflects uncertainty about the correct setup and integration limits.

PsychonautQQ
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Homework Statement


Find the area inside r = 9sinθ but outside r = 2




Homework Equations


Area = 1/2(Integral of (f(θ)^2 - g(θ)^2)dθ



The Attempt at a Solution


f(θ)^2 =
81sin^2θ = 81((1-cos(2θ))/2)
g(θ)^2 = 4

f(θ)^2 - g(θ)^2 = 36.5 - cos(2θ)/2
integral of (36.5 - cos(2θ)/2)
[36.5θ - sin(2θ)/4]

Area = 1/2[36.5θ - sin(2θ)/4]
If I integrate from 0 to ∏/2 then multiply that area times two I get 114.6681319
Which is the wrong answer.
Where did I go wrong ;-(
 
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How did you get the limits 0 to π/2? Have you sketched the two circles?
 
Should I integrate from the intersection point to pi/2?
9sin(theta) = 2
theta = arcsin(2/9)
theta = .224093
If I do that that is half of the area and then I can multiply that by two to get the full area. Does that look better to you?
 
PsychonautQQ said:
Should I integrate from the intersection point to pi/2?
Yes, that is valid because the shaded region is symmetric wrt the y axis.
9sin(theta) = 2
theta = arcsin(2/9)
theta = .224093
I would suggest subbing in arcsin(2/9). The range of this function is [-pi/2,pi/2] so subbing this in will give you the required theta.

If I do that that is half of the area and then I can multiply that by two to get the full area. Does that look better to you?
Yes.
 
CAF123 said:
I would suggest subbing in arcsin(2/9). The range of this function is [-pi/2,pi/2] so subbing this in will give you the required theta.


Yes.
I plug these in for theta and then get a required theta? I'm confused by what you mean here. Do you mean that the limits of integration are [-∏/2,∏/2]?
 
PsychonautQQ said:
I plug these in for theta and then get a required theta? I'm confused by what you mean here. Do you mean that the limits of integration are [-∏/2,∏/2]?
No, just sub in arcsin(2/9) instead of subbing in 0.224... to avoid rounding errors. The principal solution of arcsin(2/9) is the one that appears in [-pi/2,pi/2] which is the theta that the two curves intersect at in the first quadrant.
You could find the other theta corresponding to the other intersection and integrate between these two values, but your method is also good, perhaps more elegant given you have noticed the symmetry.
 

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