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

In summary, the homework statement is asking for the area of a region that lies inside the curve r^2=8cos(2θ) and outside r=2. The equation for the area is .5∫R^2(outside)-r^2(inside) dθ. The Attempt at a Solution is that r^2=8cos(2θ) and r=2, so 4=8cos(2θ) and .5=cos(2θ). Since .5 is positive, we need the angles in the first and fourth quadrant. The attempt was to solve for the cosine of the two angles, but when i plug in the values, i can't seem to
  • #1
steel1
16
0

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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  • #2
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:
  • #3
not sure what your trying to get at. :<
 
  • #4
What answer should you be getting?
 
  • #5
2.739
 
  • #6
im so dumb
 
Last edited:
  • #7
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?
 
  • #8
yes, i was getting exactly half the right answer. forgot about the symmetry, thanks!
 

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