Chapter 9: Q. 7. (page 755)

Why do we require that $0 \leq β - α \leq 2 π$ in the statement
of Theorem $9.13$ ?

Short Answer

Expert verified

According to the definition, to integrate the area $β > α$

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

Step by step solution

Given information

Consider the following theorem statement:

Let $α, β$ is any real numbers, such that $0 \leq β - α \leq 2 π \cdot R$

R is the polar plane region enclosed by the rays $θ = α$ and $θ = β$

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

According to the given information,

A continuous function $r = f (θ)$ then the area of the region $R$ is $\frac{1}{2} \int_{α}^{β} (f (θ))^{2} d θ$

To integrate the area $β > α$ according to the description

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

That is why $β - α < 2 π$

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

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Why do we require that $0 \leq β - α \leq 2 π$ in the statement
of Theorem $9.13$ ?

Short Answer

Step by step solution

Given information

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

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Why do we require that 0≤β-α≤2πin the statementof Theorem 9.13?

Short Answer

Step by step solution

Given information

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

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Decision Maths

Mechanics Maths

Geometry

Theoretical and Mathematical Physics

Logic and Functions

Discrete Mathematics

Study anywhere. Anytime. Across all devices.

Why do we require that $0 \leq β - α \leq 2 π$ in the statement
of Theorem $9.13$ ?