# Finding arc length of polar Curve

## 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?

SteamKing
Staff Emeritus
Homework Helper
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.

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

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
Gold Member
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.

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
Gold Member
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.