Q. 30 The region inside one loop of th... [FREE SOLUTION]

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Chapter 13: Q. 30 (page 1027)

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

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

$r^{2} = \sin 2 胃$

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 $r^{2} = \sin 2 胃$

The first quadrant of lemniscates

The area of the region in the first quadrant surrounded by one loop of lemniscates can be represented as $A = {鈭�}_{0}^{蟺 / 2} 鈥� {鈭�}_{0}^{r + \sqrt{s in 鈦� 2 胃}} 鈥� 谤诲谤诲胃$

Integrate first with regard to r

$\begin{aligned} A = {鈭�}_{0}^{蟺 / 2} 鈥� {[\frac{r^{2}}{2}]}_{0}^{\sqrt{\sin 鈦� 2 胃}} 诲胃 \\ A = {鈭�}_{0}^{蟺 / 2} 鈥� [\frac{\sin 鈦� 2 胃鈭� 0}{2}] 诲胃 \\ A = {[鈭� \frac{\cos 鈦� 2 胃}{4}]}_{0}^{蟺 / 2} \end{aligned}$

substituting the limits,

$\begin{aligned} A = [鈭� (\frac{\cos 鈦� 蟺鈭� \cos 鈦� 0}{4})] \\ A = [鈭� (\frac{鈭� 1 鈭� 1}{4})] \\ A = \frac{1}{2} \end{aligned}$

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

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91影视

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

Short Answer

Step by step solution

Given information

Assessing the integral

The first quadrant of lemniscates

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91影视

The region inside one loop of the lemniscater2=sin2胃

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

Decision Maths

Geometry

Probability and Statistics

Pure Maths

Discrete Mathematics

Mechanics Maths

Study anywhere. Anytime. Across all devices.

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