Finding arc length of polar Curve

[PsychonautQQ]

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

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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?

 

[SteamKing]

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.

 

[Enigman]

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

 

[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) ?

 

[CAF123]

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.

 

[PsychonautQQ]

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∏?

 

[CAF123]

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.