Area between Polar curves

  • #1
PsychonautQQ
784
11

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 ;-(
 

Answers and Replies

  • #2
CAF123
Gold Member
2,950
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How did you get the limits 0 to π/2? Have you sketched the two circles?
 
  • #3
PsychonautQQ
784
11
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?
 
  • #4
CAF123
Gold Member
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88
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.
 
  • #5
PsychonautQQ
784
11
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]?
 
  • #6
CAF123
Gold Member
2,950
88
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.
 

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