Chapter 13: Q. 24 (page 1027)

Each of the integrals or integral expressions in Exercises represents the area of a region in the plane. Use polar coordinates to sketch the region and evaluate the expressions.
$2 \int_{0}^{\frac{π}{2}} \int_{0}^{\sin θ} rdrdθ$

Short Answer

Expert verified

The integral value is $2 \int_{0}^{\frac{π}{2}} \int_{0}^{\sin θ} rdrdθ = \frac{π}{4}$

Step by step solution

Given information

The integral function is $2 \int_{0}^{\frac{π}{2}} \int_{0}^{\sin θ} rdrdθ$

Calculation of derivatives

The goal of this issue is to sketch the region and assess the expression using polar coordinates $2 \int_{0}^{\frac{π}{2}} \int_{0}^{\sin θ} rdrdθ$

using the values , $r = 0, r = \sin θ and θ = 0, θ = π / 2$

$θ$	$r = \sin θ$ localid="1652361865068" $r = sinθ$
0	0
localid="1652361870625" $π / 6$	0.5
localid="1652361873973" $π / 4$	0.7071
localid="1652361878453" $π / 3$	0.8660
localid="1652361882858" $π / 2$	1.0

Use the table above to draw the region ,

elaborating the expressions,

localid="1652361900299" $\begin{aligned} 2 \int_{0}^{π / 2} \int_{0}^{\sin 3 θ} rdrdθ = 2 \int_{0}^{π / 2} {[\frac{r^{2}}{2}]}_{0}^{\sin 3 θ} dθ \\ = 2 \int_{0}^{π / 2} [\frac{\sin^{2} 3 θ - 0}{2}] dθ \\ = 2 \int_{0}^{π / 2} [\frac{(1 - \cos 6 θ)}{4}] dθ \end{aligned}$

We can integrating in terms of localid="1652361916107" $θ,$

localid="1652361908352" $2 \int_{0}^{π / 2} \int_{0}^{\sin 3 θ} rdrdθ = 2 {[\frac{{θ - (\sin 6 θ) / 6}}{4}]}_{0}^{π / 2} [\int \cos xdx = \sin x]$

Substituting the limits of derivatives,

localid="1652361934374" $2 \int_{0}^{π / 2} \int_{0}^{\sin 3 θ} rdrdθ = 2 [\frac{{\frac{π / 2 - (\sin 3 π)}{6} - 0}}{4}] = \frac{π}{4}$

As a result, the integral value is localid="1652361944670" $2 \int_{0}^{π / 2} \int_{0}^{\sin 3 θ} rdrdθ = \frac{π}{4}$

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Each of the integrals or integral expressions in Exercises represents the area of a region in the plane. Use polar coordinates to sketch the region and evaluate the expressions.
$2 \int_{0}^{\frac{π}{2}} \int_{0}^{\sin θ} rdrdθ$

Short Answer

Step by step solution

Given information

Calculation of derivatives

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Each of the integrals or integral expressions in Exercises represents the area of a region in the plane. Use polar coordinates to sketch the region and evaluate the expressions. 2∫0π2 ∫0sin⁡θ rdrdθ

Short Answer

Step by step solution

Given information

Calculation of derivatives

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Logic and Functions

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Theoretical and Mathematical Physics

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Study anywhere. Anytime. Across all devices.

Each of the integrals or integral expressions in Exercises represents the area of a region in the plane. Use polar coordinates to sketch the region and evaluate the expressions.
$2 \int_{0}^{\frac{π}{2}} \int_{0}^{\sin θ} rdrdθ$