Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How to calculate the area between 2 polar curves: $r=2-\sin\theta$ and $r=3\sin\theta$?

I know that one curve is a limaçon and the other is a circle. I have them drawn out as well, my only question is once I've found the points of intersection, how do I know which way to integrate from and which to subtract/add from the other?

share|improve this question
1  
From $0$ to $\pi/6$ the first curve we meet is the circle. Then it is the other guy up to $\pi/2$ (and beyond). But there is symmetry, so go to $\pi/2$ only and double. –  André Nicolas Apr 9 '13 at 6:07
    
If you have them figured out with intersection points and everything, I guess you just have to integrate $1$ in polar coordinates over the region $A$ they bound: $\iint _A 1 rdrd\theta$. Then use what you know about $A$ to do Fubini appropriately. –  1015 Apr 9 '13 at 6:07
1  
For each part integrate $\frac{1}{2}r^2\,d\theta$. Since we want to double, forget about the $\frac{1}{2}$ and forget about doubling. We get $\int_0^{\pi/6} 9\cos^2\theta\,d\theta$ plus $\int_{\pi/6}^{\pi/2}(2-\sin\theta)^2\,d\theta$. –  André Nicolas Apr 9 '13 at 6:12
    
thanks @julien ! –  lawlipop Apr 9 '13 at 6:14
1  
What is this "4/2"? –  zyx Apr 9 '13 at 6:26
show 2 more comments

1 Answer

up vote 3 down vote accepted

Make a careful sketch. Or have software do it for you. We want the area that is common to the regions enclosed by the two curves.

The two curves meet at $\theta=\pi/6$ and $\theta=\pi-\pi/6$.

Looking outward from the origin, from $\theta=0$ to $\theta=\pi/6$, the first curve we meet is the circle. Then from $\pi/6$ to $\pi/2$ (and beyond), it's the other guy up to $5\pi/6$. Then it is the circle again up to $\pi$. But there is symmetry, so go to $\pi/2$ only and double.

For each part integrate $\frac{1}{2}r^2\,d\theta$. Since we intended to double, forget about the $\frac{1}{2}$ and forget about doubling. We get $$\int_0^{\pi/6} 9\sin^2\theta\,d\theta + \int_{\pi/6}^{\pi/2}(2-\sin\theta)^2\,d\theta.$$

share|improve this answer
    
I did not draw these curves, which is maybe why I don't understand. But shouldn't the integrands be $\pm(9\sin^2\theta-(2-\sin\theta)^2)$ to get the area between the curves? –  1015 Apr 9 '13 at 12:46
    
@julien: Unfortunately, I do not understand the suggestion. But it is early in the morning, pre-coffee. The idea is that we draw angular sectors of angular width $d\theta$ from the origin to the curve, and add up the infinitesimal areas $\frac{1}{2}r^2\,d\theta$. The answer writes itself if we have the curves in front of our eyes. –  André Nicolas Apr 9 '13 at 13:41
    
For instance: consider the polar curves $r_1(\theta)=1$ and $r_2(\theta)=2$ (two circles, of course). The area between them is $3\pi=\pi 2^2-\pi 1^2=\frac{1}{2}\int_0^{2\pi}(r_2^2(\theta)-r_1^2(\theta)d\theta$. And it is not $\frac{1}{2}\int_0^{2\pi}(r_2^2(\theta)d\theta=4\pi$ as your approach seems to suggest, keeping only the curve which is further away from the origin. –  1015 Apr 9 '13 at 13:47
    
@julien: I was not suggesting a universal algorithm. In our case, if we are at the origin and stare outwards, every point of our region is visible. But the functional form of the outer boundary varies. A picture reveals all. –  André Nicolas Apr 9 '13 at 13:56
    
Sorry. I should not speak without actually drawing the picture... –  1015 Apr 9 '13 at 13:57
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.