# How to calculate the area between 2 polar curves: $r=\frac{4}{2}-\sin\theta$ and $r=3\sin\theta$?

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?

-
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
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
What is this "4/2"? – zyx Apr 9 '13 at 6:26

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 $$\int_0^{\pi/6} 9\sin^2\theta\,d\theta + \int_{\pi/6}^{\pi/2}(2-\sin\theta)^2\,d\theta.$$

-
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