What is the Area Between Two Polar Curves?

  • Thread starter Thread starter PsychonautQQ
  • Start date Start date
  • Tags Tags
    Area Curves Polar
Click For Summary
SUMMARY

The discussion focuses on calculating the area between two polar curves defined by the equations r = 9sin(θ) and r = 2. The area is determined using the formula Area = 1/2(Integral of (f(θ)^2 - g(θ)^2)dθ). The correct limits of integration are established as the intersection points of the curves, specifically using θ = arcsin(2/9) to avoid rounding errors. The final area calculation involves integrating from this intersection point to π/2 and then multiplying by two due to the symmetry of the shaded region.

PREREQUISITES
  • Understanding of polar coordinates and curves
  • Familiarity with integration techniques in calculus
  • Knowledge of trigonometric functions and their properties
  • Ability to apply the area formula for polar curves
NEXT STEPS
  • Study the derivation of the area formula for polar curves
  • Learn how to find intersection points of polar equations
  • Practice integrating trigonometric functions over specified intervals
  • Explore symmetry in polar graphs and its implications for area calculations
USEFUL FOR

Students studying calculus, particularly those focusing on polar coordinates, as well as educators and tutors looking for examples of area calculations between polar curves.

PsychonautQQ
Messages
781
Reaction score
10

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 ;-(
 
Physics news on Phys.org
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.
 

Similar threads

Replies
7
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
4K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K