Chapter 13: Q. 30 (page 1027)

The region inside one loop of the lemniscate $r^{2} = \sin 2 θ$

Short Answer

Expert verified

In the first quadrant, the region bordered by one loop of lemniscates is $A = \frac{1}{2}$

Step by step solution

Given information

$r^{2} = \sin 2 θ$

Assessing the integral

The goal of this issue is to find and evaluate an iterated integral in polar coordinates that reflects the area of a given region in the polar plane.

The lemniscates encircle the region $r^{2} = \sin 2 θ$

The first quadrant of lemniscates

The area of the region in the first quadrant surrounded by one loop of lemniscates can be represented as $A = \int_{0}^{π / 2} \int_{0}^{r + \sqrt{s in 2 θ}} rdrdθ$

Integrate first with regard to r

$\begin{aligned} A = \int_{0}^{π / 2} {[\frac{r^{2}}{2}]}_{0}^{\sqrt{\sin 2 θ}} dθ \\ A = \int_{0}^{π / 2} [\frac{\sin 2 θ - 0}{2}] dθ \\ A = {[- \frac{\cos 2 θ}{4}]}_{0}^{π / 2} \end{aligned}$

substituting the limits,

$\begin{aligned} A = [- (\frac{\cos π - \cos 0}{4})] \\ A = [- (\frac{- 1 - 1}{4})] \\ A = \frac{1}{2} \end{aligned}$

The area of the region in the first quadrant limited by one loop of lemniscates is $A = \frac{1}{2}$

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The region inside one loop of the lemniscate $r^{2} = \sin 2 θ$

Short Answer

Step by step solution

Given information

Assessing the integral

The first quadrant of lemniscates

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The region inside one loop of the lemniscater2=sin2θ

Short Answer

Step by step solution

Given information

Assessing the integral

The first quadrant of lemniscates

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

Calculus

Applied Mathematics

Logic and Functions

Decision Maths

Statistics

Study anywhere. Anytime. Across all devices.

The region inside one loop of the lemniscate $r^{2} = \sin 2 θ$