# Homework Help: Find the Exact length of the Polar Curve

1. Apr 29, 2012

### CitizenInsane

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

L= ∫√(r^2+(dr/dθ)^2)dθ

3. 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.

Last edited: Apr 29, 2012
2. Apr 29, 2012

### HallsofIvy

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!

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 $2\pi$.

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

3. Apr 29, 2012

### sharks

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.

#### Attached Files:

• ###### polar.gif
File size:
4.8 KB
Views:
461
4. Apr 29, 2012

### 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.

5. Apr 29, 2012

### sharks

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?

Last edited: Apr 29, 2012