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!
 

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

1. What is a polar curve area question?

A polar curve area question is a mathematical problem in which the area enclosed by a polar curve is calculated using integration techniques.

2. How is a polar curve area question solved?

A polar curve area question is solved by first converting the polar equation into rectangular form, then using integration to find the area under the curve.

3. What is the difference between a polar curve area question and a regular area question?

The main difference between a polar curve area question and a regular area question is that the polar curve area question involves converting the polar equation into rectangular form, while a regular area question can be solved using basic geometric formulas.

4. Can a polar curve area question have negative solutions?

No, a polar curve area question cannot have negative solutions since area is always a positive value.

5. How can a polar curve area question be applied in real life?

Polar curve area questions can be applied in various fields of science and engineering, such as physics, astronomy, and architecture. They can be used to calculate the area of irregular shapes or to determine the volume of objects with curved surfaces.

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