# Polar coords

by daster
Tags: coords, polar
 P: n/a 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.
 HW Helper P: 2,885 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.
 P: n/a Why do we want the value of theta at r=0?
HW Helper
P: 2,885

## Polar coords

 Quote by daster Why do we want the value of theta at r=0?
Have you even sketched the graph yet ? It will become clear as day if you do.
 P: n/a 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. So I'm just wondering if my method was correct.
 HW Helper P: 2,885 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 ?
 P: n/a 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. Anyway, thanks for your help. I appreciate it.
 HW Helper P: 2,885 Sure.

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