# Polar coords

1. Jan 18, 2005

### daster

I need help with finding areas. I'm having trouble picking the correct limits for my integration.

Say for example we had r=2cos2t and a half-line t=pi/6, and I want to find the small area bounded between them.

$$\frac{1}{2}\int (2\cos (2\theta))^{2}\,d\theta$$

I can do the integration, but what do I choose as its limits?

I'm not particularly interested in the answer to this question; I'm looking for a good explanation or maybe a couple of pointers.

2. Jan 18, 2005

### Curious3141

At what value(s) of theta does $r$ become zero ? Set the r(theta) equation to zero and solve for theta. Use the 1st quadrant value of theta obtained as the upper bound with $\frac{\pi}{6}$ as the lower.

3. Jan 18, 2005

### daster

Why do we want the value of theta at r=0?

4. Jan 18, 2005

### Curious3141

Have you even sketched the graph yet ? It will become clear as day if you do.

5. Jan 18, 2005

### daster

Can you please check my working?

(a) Sketch the curve with polar equation

$$r=3\cos 2\theta, \, -\frac{\pi}{4}\leq\theta<\frac{\pi}{4}$$.

The curve looks like 1 rose petal.

(b) Find the area of the smaller finite region enclosed between the curve and the half-line $\theta=\frac{\pi}{6}$.

$$Area = \frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} (3\cos 2\theta)^2\,d\theta = \frac{9}{4}\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} (1+\cos 4\theta)\,d\theta = \frac{9}{4}\left[\theta+\frac{1}{4}\sin 4\theta\right]_{\frac{\pi}{6}}^{\frac{\pi}{4}} = \frac{3\pi}{16}- \frac{9\sqrt{3}}{32}$$

A couple of arithmetic slips probably made it through, since I did all of this using my keyboard & notepad.exe. :tongue2: So I'm just wondering if my method was correct.

Last edited by a moderator: Jan 18, 2005
6. Jan 18, 2005

### Curious3141

The coefficient is 3 ? Not 2 ? If that's the case then your working in the last post is correct. But why did you give a different equation in your first post ?

7. Jan 18, 2005

### daster

In my first post I made up an equation and a line, and when I went to do a question from my textbook, it happened to be almost like the one I made up. :tongue2:

Anyway, thanks for your help. I appreciate it.

8. Jan 18, 2005

Sure.