# Polar Integration of $r = 2\cos(\theta)$

$r = 2\cos(\theta)$ has the graph

I want to know why the following integral to find area does not work $$\int_0^{2 \pi } \frac{1}{2} (2 \cos (\theta ))^2 \, d\theta$$

whereas this one does:

$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2} } \frac{1}{2} (2 \cos (\theta ))^2 \, d\theta$$

Why do the limits of integration have to go from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$? Doesn't going from $0$ to $2\pi$ also sweep out the full circle?

the limit is from $0$ to $\pi$ not $0$ to $2\pi$ if you do the latter, you will go around the circle twice.

at $\theta = 0$ you are $(2,0)$ at $\theta = \pi/3$ you are at $x = 1, y = 1$ at $\theta = \pi/2$ you are at $(0,0)$ and then you move in the lower semicircle for the next $\pi/2$ to $\pi$ if you go farther in $\theta$ you will go back to where you started.

one has to be careful when setting up polar integral about the limits and how the curve is traced. you will lots of one inside other in cardiods and more complicated figures like $r = \cos(5 \theta),$ etc

The families of polar curves $\ r \ = \ a \ \cos \ n \theta \$ and $\ r \ = \ a \ \sin \ n\theta \$ are called "rosettes", because they form symmetrical "flower-petal" arrangements. Those with $\ n \$ even trace their respective curves once in a period of $\ 2 \pi \$ and have $\ 2n \$ "petals", while those with $\ n \$ odd trace their curves, which have $\ n \$ "petals", in a cycle of length $\ \pi \$ and will then retrace the curve in subsequent periods of $\ \pi \$ . (For these latter rosettes, the "negative radius" portions of the curves "fall on top of" the "positive radius" portions.)

The curves $\ r \ = \ a \ \cos \ \theta \$ and $\ r \ = \ a \ \sin \ \theta \$ are certainly circular (as transforming them to Cartesian coordinates will confirm), but in fact belong to the families of rosettes. They are "one-petal" rosettes, since $\ n \ = \ 1 \$ . So these curves are among the ones retraced by running through an angular interval of length $\ 2 \pi \$ . (This is why the integrations for these curves are "stopped" after an interval of $\ \pi \$ radians.)

Start with $\theta=0$ you get $r=2$ then move to $\theta=\frac{\pi}{2}$ you will get $r=0$ (then you have from here the upper half of the circle). From $\frac{\pi}{2}$ to $\pi$ you have a negative $r$ which don't have really much sense. Then you have to consider it from $\frac{3\pi}{2}$ to $2\pi$ to complete the lower half.

$$\int_{0}^{\frac{\pi}{2} } \frac{1}{2} (2 \cos (\theta ))^2 \, d\theta+\int_{\frac{3\pi}{2}}^{2\pi} \frac{1}{2} (2 \cos (\theta ))^2 \, d\theta=\int_{0}^{\frac{\pi}{2} } \frac{1}{2} (2 \cos (\theta ))^2 \, d\theta+\int_{-\frac{\pi}{2}}^{0} \frac{1}{2} (2 \cos (\theta ))^2 \, d\theta$$ $$=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{2} (2 \cos (\theta ))^2 \, d\theta$$

• Actually I think you are overcomplicating it. The negative $r$ value means that the ray is in the opposite direction. You don't need to split up the integral at all. The reason why mine gave $2\pi$ was because I was double counting the area as @abel explained quite nicely below. – 1110101001 Jan 14 '15 at 2:00