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

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SUMMARY

The area of the region inside the polar curve defined by r² = 8cos(2θ) and outside the circle r = 2 can be calculated using the formula for the area of polar curves: 0.5∫(R² - r²) dθ. The critical angles for integration are θ = π/6 and θ = -π/6, derived from setting r² = 4. The final area calculation must account for symmetry, as the area on the left side of the y-axis is identical to that on the right, resulting in a total area of 2.739.

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  • Knowledge of trigonometric identities, specifically for cos(2θ)
  • Ability to evaluate definite integrals
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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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