Find the Exact length of the Polar Curve

[CitizenInsane]

Homework Statement

Find the Exact length of the Polar Curve for
r=2(1+cosθ)

No limits of Integration were given which I found to be odd.

Homework Equations

L= ∫√(r^2+(dr/dθ)^2)dθ

The Attempt at a Solution

r=2(1+cosθ)
dr/dθ=-2sinθ

L=∫√((2+2cosθ)^2+(-2sinθ)^2)dθ
=∫√(4cos^2θ+4sin^2θ+8cosθ+4)dθ
=∫√(8cosθ+8)dθ

Got to this point and figured I did something incorrectly.

 

[HallsofIvy]

Homework Statement

r=2(1+cosθ)

It's not a good idea to put the statement of the problem only in the title! I started to say "no that is not a problem statement" until I looked up at your title!

No limits of Integration were given which I found to be odd.

Perhaps they expect you to be able to get that yourself. When \theta= 0, cos(\theta)= 1 so r= 4; when \theta= \pi/2, cos(\theta)= 0 so r= 2; when \theta= pi, cos(\theta)= -1 so r= 0; when \theta= 3\pi/2, cos(\theta)= 0 so r= 2 and when \theta= 2\pi, cos(\theta)= 1 so r= 4 again. We make one complete loop around the figure (a "cardiod" since it looks roughly like the "valentine" heart) as \theta goes from 0 to 2\pi. That shouldn't be too surprizing since cosine has period 2\pi.

Homework Equations

L= ∫√(r^2+(dr/dθ)^2)dθ

The Attempt at a Solution

r=2(1+cosθ)
dr/dθ=-2sinθ

L=∫√((2+2cosθ)^2+(-2sinθ)^2)dθ
=∫√(4cos^2θ+4sin^2θ+8cosθ+4)dθ
=∫√(8cosθ+8)dθ

Got to this point and figured I did something incorrectly.

No, that's correct. Now integrate from 0 to 2\pi.

 

[DryRun]

Like HallsofIvy said, you should plot the polar curve: r=2(1+cosθ) to find the limits of integration. I have attached the graph to this post.

 

[CitizenInsane]

Sorry for not putting enough info originally, new to the forums.
Also when I integrate, I get stuck at
=√8∫√(cosθ+1)dθ
Only step I could think of next is to u-sub what's inside the radical.

 

[DryRun]

Sorry for not putting enough info originally, new to the forums.
Also when I integrate, I get stuck at
=√8∫√(cosθ+1)dθ
Only step I could think of next is to u-sub what's inside the radical.

Try trigonometric substitution. Let \sqrt{\cos \theta}=\tan \phi

After integrating, the answer should be: 2\sqrt{8}.\sqrt{\cos \theta +1}.\tan \frac{\theta}{2} Now, you have to put the limits and evaluate.
The problem is that for 0 \leq \theta \leq 2\pi, the evaluation gives 0.

Are you sure the integral for L is correct in your first post?