# Finding arc length of polar Curve

1. Sep 9, 2013

### PsychonautQQ

1. The problem statement, all variables and given/known data
Find the arc length of polar curve 9+9cosθ

2. Relevant equations

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

)3. 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?

2. Sep 9, 2013

### SteamKing

Staff Emeritus
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. Sep 9, 2013

### Enigman

For the other thing please use LaTeX...just a suggestion.

4. Sep 9, 2013

### PsychonautQQ

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. Sep 9, 2013

### CAF123

Yes, a half angle sub will help to evaluate it.

6. Sep 9, 2013

### PsychonautQQ

Why use a half angle sub? Doesn't it just come out to 162θ+162sinθ between 0 and 2∏?

7. Sep 10, 2013

### CAF123

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.