What is the Area of the Region Inside a Polar Curve and Outside a Given Circle?

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Homework Help Overview

The discussion revolves around finding the area of the region that lies inside the polar curve defined by r² = 8cos(2θ) and outside the circle defined by r = 2. The problem involves concepts from polar coordinates and area calculation in calculus.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the setup of the problem, including the equations for the curves involved. There are attempts to evaluate the integral for the area, with some participants questioning their calculations and the implications of symmetry in the area.

Discussion Status

The discussion is ongoing, with participants sharing their attempts and questioning their understanding of the symmetry in the problem. Some guidance has been provided regarding the consideration of symmetric areas, but no consensus has been reached on the final solution.

Contextual Notes

Some participants express confusion about the calculations and the interpretation of the polar curves, particularly regarding the angles and the symmetry of the area being calculated.

steel1
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Homework Statement


Find the area of the region that lies inside the curve r^2=8cos(2θ) and outside r=2

Homework Equations


area of polar curves = .5∫R^2(outside)-r^2(inside) dθ

The Attempt at a Solution


r^2=8cos(2θ) and r=2, so...

4=8cos(2θ)
.5=cos(2θ) since .5 is positive, we need the angles in the first and fourth quadrant
2θ=∏/3 and -∏/3
θ=∏/6 and -∏/6

.5∫8cos(2θ)-4 from -pi/6 to pi/6
=difference of .5[4sin(2θ)-4θ] evaluated at pi/6 and -pi/6. but when i plug the values in, i can't seem to get the right answer
 
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steel1 said:

Homework Statement


Find the area of the region that lies inside the curve r^2=8cos(2θ) and outside r=2

Homework Equations


area of polar curves = .5∫R^2(outside)-r^2(inside) dθ

The Attempt at a Solution


r^2=8cos(2θ) and r=2, so...

4=8cos(2θ)
.5=cos(2θ) since .5 is positive, we need the angles in the first and fourth quadrant
2θ=∏/3 and -∏/3
θ=∏/6 and -∏/6

.5∫8cos(2θ)-4 from -pi/6 to pi/6
=difference of .5[4sin(2θ)-4θ] evaluated at pi/6 and -pi/6. but when i plug the values in, i can't seem to get the right answer
Consider what happens when cos(2θ) < 0 .

Ignore this post. I misread the problem. DUH !
 
Last edited:
not sure what your trying to get at. :<
 
What answer should you be getting?
 
2.739
 
im so dumb
 
Last edited:
steel1 said:

Homework Statement


Find the area of the region that lies inside the curve r^2=8cos(2θ) and outside r=2


Homework Equations


area of polar curves = .5∫R^2(outside)-r^2(inside) dθ


The Attempt at a Solution


r^2=8cos(2θ) and r=2, so...

4=8cos(2θ)
.5=cos(2θ) since .5 is positive, we need the angles in the first and fourth quadrant
2θ=∏/3 and -∏/3
θ=∏/6 and -∏/6

.5∫8cos(2θ)-4 from -pi/6 to pi/6
=difference of .5[4sin(2θ)-4θ] evaluated at pi/6 and -pi/6. but when i plug the values in, i can't seem to get the right answer

Are you getting exactly half the desired answer? Are you aware there is a symmetric area on the left side of the y axis?
 
yes, i was getting exactly half the right answer. forgot about the symmetry, thanks!
 

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