# Area under Polar Curves: Where did I go wrong?

1. Sep 4, 2009

### CheapFungus

1.Where did I go wrong in finding the area enclosed inside r = 3 cos θ?

2. Relevant equations

I used the formula 1/2 ∫ ((f(θ)) squared dθ from alpha to beta

3. The attempt at a solution

I looked for the area of the semicircle from 0 to pi and then multiplied the whole thing by 2, since the equation is a circle, so it looks like this:

=>2*1/2 ∫ from 0 to pi (3 cos θ)squared dθ
=> ∫ from 0 to pi (9 cos squared θ) dθ
=> ∫ from 0 to pi (9 (1/2) (1+cos 2θ) dθ [Using half angle identities]
=> 9/2 ∫ from 0 to pi (1+cos 2θ) dθ
Integrating I get :
=> 9/2 (θ + 1/2 sin 2θ ] from 0 to pi)
=> 9/2 [(pi + 1/2 sin 2pi) - (0 + 1/2 sin 2o)]
=> 9/2 [pi + 0 - 0-0]

So I get the final answer of 9/2pi. Thing is, the answer in the book is 9/4pi and I can't figure out they go there. Help, anyone?

Last edited: Sep 5, 2009
2. Sep 4, 2009

### Dick

If you trace the curve carefully from theta=0 to pi, you'll realize it's not a semicircle. It's the whole circle. r is negative for theta between pi/2 and pi.

3. Sep 5, 2009

### CheapFungus

Sorry, I phrased that wrong. I took the area of the half circle first, then multiplied it by two.

Anyway, I am quite confused by the limits. How do you get the limits? >_<

Last edited: Sep 5, 2009
4. Sep 5, 2009

### Dick

You get the limits by sketching the curve. It's starts at (3,0) (in xy coordinates) at theta=0, r=3 and returns to (3,0) at theta=pi, r=(-3). Calling that either a semicircle or a half circle are BOTH wrong. Don't multiply by 2!

5. Sep 5, 2009

### CheapFungus

I see! Is there a general way for finding the limits? It seems that I always get messed up in that part...

6. Sep 5, 2009

### Dick

Just what I've already suggested. You'll have to learn to look at enough points on the curve to draw a good sketch.