#1
Jan1805, 06:34 PM

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 halfline t=pi/6, and I want to find the small area bounded between them. [tex]\frac{1}{2}\int (2\cos (2\theta))^{2}\,d\theta[/tex] 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
Jan1805, 07:45 PM

HW Helper
P: 2,885

At what value(s) of theta does [itex]r[/itex] 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 [itex]\frac{\pi}{6}[/itex] as the lower.



#3
Jan1805, 08:11 PM

P: n/a

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




#4
Jan1805, 08:14 PM

HW Helper
P: 2,885

Polar coords 


#5
Jan1805, 09:17 PM

P: n/a

Can you please check my working?
(a) Sketch the curve with polar equation [tex]r=3\cos 2\theta, \, \frac{\pi}{4}\leq\theta<\frac{\pi}{4}[/tex]. The curve looks like 1 rose petal. (b) Find the area of the smaller finite region enclosed between the curve and the halfline [itex]\theta=\frac{\pi}{6}[/itex]. [tex]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}[/tex] 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. 



#6
Jan1805, 09:26 PM

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 ?



#7
Jan1805, 09:30 PM

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. 


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