# Thread: Finding Polar Limits of Integration

1. ## Finding Polar Limits of Integration

I'm currently reading an AP Calculus BC review book, and one of its example problems has to do with finding the area of the inner loop of a limaçon.

The formula for the limaçon is $r=1+2\cos\theta$. I'm supposed to find the area of just its inner loop. The book begins its problem walkthrough by stating that "in the inner loop, $r$ sweeps out the region as $\theta$ goes from $\frac{2\pi}{3}$ to $\frac{4\pi}{3}$." Those would, therefore, be the limits of integration.

I'm really shaky on polar coordinate geometry, though, so I have no idea how you're supposed to have figured that out. How do you find the $\theta$ values on a polar coordinate graph? I'm familiar with the basic polar-cartesian conversion equations $x=r\cos\theta$, $y=r\sin\theta$, $r=\sqrt{x^2+y^2}$, and $\theta=\arctan{\frac{y}{x}}$, but I don't know how to find $\theta$ values when the polar graph is as complicated as a looped limaçon.

2. Originally Posted by greenstupor
I'm currently reading an AP Calculus BC review book, and one of its example problems has to do with finding the area of the inner loop of a limaçon.

The formula for the limaçon is $r=1+2\cos\theta$. I'm supposed to find the area of just its inner loop. The book begins its problem walkthrough by stating that "in the inner loop, $r$ sweeps out the region as $\theta$ goes from $\frac{2\pi}{3}$ to $\frac{4\pi}{3}$." Those would, therefore, be the limits of integration.

I'm really shaky on polar coordinate geometry, though, so I have no idea how you're supposed to have figured that out. How do you find the $\theta$ values on a polar coordinate graph? I'm familiar with the basic polar-cartesian conversion equations $x=r\cos\theta$, $y=r\sin\theta$, $r=\sqrt{x^2+y^2}$, and $\theta=\arctan{\frac{y}{x}}$, but I don't know how to find $\theta$ values when the polar graph is as complicated as a looped limaçon.

To get to the inner loop of the limaçon, the idea is to find values of $\theta$ that cause $r$ to be negative!

So solve the inequality $1+2\cos\theta<0\implies\cos\theta<-\tfrac{1}{2}$. It turns out in this case that $\frac{2\pi}{3}< \theta<\frac{4\pi}{3}$ does the job.

Does this help?