Submitted to HEJ Manuscript no.: ANM-010201-A

Areas in polar coordinates

For many purposes it is important to be able to calculate areas using polar coordinates. Let $r=f(\theta)$ be the equation of a curve in polar coordinates. Let $I$ be the area of the region which is bounded by the $x$-axis (that is, the line $\theta =0$), the line through the origin making an angle $\theta$ with the $x$-axis, and the portion of the curve between these two lines. By the fundamental theorem of the integral calculus, the area of the sector between the polar angles $\alpha$ and $\beta$ is given by the expression

$$\frac{1}{2}\,\int _{\alpha }^{\beta }{\rm f}(\theta )^{2}\,d\theta .$$

If $\beta>\alpha$, this expression cannot be less than zero.

EXAMPLE 7.2.

Consider the area bounded by the one loop of a lemniscate. The equation of the lemniscate is $r^{2}=2\,a^{2}\,{\rm cos}\,2\,\theta$, and we obtain one loop by letting $\theta$ vary from $-\frac {\pi }{4}$ to $+\frac {\pi }{4}$. The shape of one loop of the lemniscate with $a=1$ is

$>$ plot([2*cos(2*theta),theta,theta=-Pi/4..Pi/4],coords=polar);

 

Figure 13:

The area of the loop in general is

$>$ a^2*Int(cos(2*theta),theta=-Pi/4..Pi/4);

$$a^{2}\,{\displaystyle \int _{ - 1/4\,\pi }^{1/4\,\pi }} {\rm cos} (2\,\theta )\,d\theta .$$

We find that the value of the integral is

$>$ value(%);

$$a^{2}$$

Submitted to HEJ Manuscript no.: ANM-010201-A