Question

Why do we require that $0\le \beta -\alpha \le 2\pi$in the statement

of Theorem $9.13$?

Short answer

According to the definition, to integrate the area $\beta >\alpha$

The area should be calculated only once for a given curve.

Step-by-step solution

Given information

Consider the following theorem statement:

Let $\alpha ,\beta$is any real numbers, such that $0\le \beta -\alpha \le 2\pi ·R$

R is the polar plane region enclosed by the rays$\theta =\alpha$and$\theta =\beta$

Explain why β-α<2π that we don't compute the portion of the area twice.

According to the given information,

A continuous function $r=f\left(\theta \right)$then the area of the region $R$is $\frac{1}{2}{\int }_{\alpha }^{\beta }\left(f\right(\theta ){)}^{2}d\theta$

To integrate the area $\beta >\alpha$according to the description

For each curve, the area should only be calculated once.

That is why $\beta -\alpha <2\pi$

So that we don't compute the portion of the area twice.