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Finding arc length of polar Curve

  • #1
784
11

Homework Statement


Find the arc length of polar curve 9+9cosθ


Homework Equations



L = integral of sqrt(r^2 + (dr/dθ)^2 dθ
dr/dθ = -9sinθ
r = 9+9cosθ

)

The Attempt at a Solution



1. Simplifying the integral
r^2 = (9+9cosθ^2) = 81 +162cosθ + 81cos^2(θ)
(dr/dθ)^2 = 81sin^2(θ)

r^2 + (dr/dθ)^2 = 81 + 162cosθ + 81cos^2(θ) + 81sin^2(θ)
81sin^2(θ) + 81cos^2(θ) = 81

162 + 162cosθ = r^2 + (dr/dθ)^2

now I have to take the integral of the squareroot...

Integral of sqrt(162 + 162cosθ)dθ
chain rule..?
(2/3)(162+162cosθ)^3/2*(162θ + 162sinθ)

Integrated between 0 and 2pi...?
which would lead to a crazy high number that I got as 3957501.966.
Anyone know where I went wrong?
 

Answers and Replies

  • #2
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,798
1,666
Well, for one thing, your indefinite integration is wrong. You can't use the power formula because you don't have the proper du outside the square root. You need to use a trig substitution instead.
 
  • #3
662
307
For the other thing please use LaTeX...just a suggestion.
 
  • #4
784
11
Sorry i'll learn LaTex soon, in a bit of a pinch at the moment.

is the final integral end up being
integral of(162 + 162cos(theta) ???
 
  • #5
CAF123
Gold Member
2,894
88
Sorry i'll learn LaTex soon, in a bit of a pinch at the moment.

is the final integral end up being
integral of(162 + 162cos(theta) ???
Yes, a half angle sub will help to evaluate it.
 
  • #6
784
11
Yes, a half angle sub will help to evaluate it.
Why use a half angle sub? Doesn't it just come out to 162θ+162sinθ between 0 and 2∏?
 
  • #7
CAF123
Gold Member
2,894
88
Why use a half angle sub? Doesn't it just come out to 162θ+162sinθ between 0 and 2∏?
Sorry,I misread what you wrote: it should have been $$\int_0^{2\pi} \sqrt{162 + 162\cos\theta}\,d\theta,$$ which you will need to use a trig sub to evaluate. Factor out the 162 from the sqrt first.
 

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