Q. 7. - Calculus

Question

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

of Theorem $9.13$?

Step-by-Step Solution

Answer

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

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

Step 1: 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 $\theta =\beta$

Step 2: Explain why β - α < 2 π   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.