Chapter 13: Q. 58 (page 1028)

Sketch the region of integration for each of integrals in Exercises 57–60, and then evaluate the integral by converting to polar coordinates.
$\int_{0}^{1} \int_{(\sqrt{3} / 3) y}^{\sqrt{4 - y^{2}}} \sqrt{x^{2} + y^{2}} d x d y$

Short Answer

Expert verified

The required value of integral is $\int_{0}^{1} \int_{(\sqrt{3} / 3) y}^{\sqrt{4 - y^{2}}} \sqrt{x^{2} + y^{2}} d x d y = \frac{4 π}{9}$

Step by step solution

Given information

The expression is $\int_{0}^{1} \int_{(\sqrt{3} / 3) y}^{\sqrt{4 - y^{2}}} \sqrt{x^{2} + y^{2}} d x d y$

Calculation

The goal of this challenge is to draw the integration zone and then calculate the integral using polar coordinates.

The foundation is

$I = \int_{0}^{1} \int_{(\sqrt{3} / 3) y}^{\sqrt{4 - y^{2}}} \sqrt{x^{2} + y^{2}} d x d y$

Here, $x = (\sqrt{3} / 3) y$ and $x = \sqrt{4 - y^{2}}, y = 0$ and $y = 1$

The region of integration R is shown in the figure.

Substitute and at the lower limit of $x$

$r \cos θ = (\sqrt{3} / 3) r \sin θ r (\cos θ - \frac{1}{\sqrt{3}} \sin θ) = 0$

This show $r = 0$ and $\tan θ = \frac{1}{\sqrt{3}}$

$\tan θ = \frac{1}{\sqrt{3}} \Rightarrow θ = \frac{π}{6}$

$x = r \cos θ$ and $y = r \sin θ$ in the upper limit of $x$ .

Substitute $y = r \sin θ$ in the lower limit of $y$

$r \sin θ = 0 \Rightarrow r = 0, θ = 0$

Thus, the limits of $r$ are $r = 0$ and $r = 2$ and that of $θ$ are 0 and $\frac{π}{6}$ .

$d x d y = r d r d θ$

Therefore,

Integrate with respect to $x$ first

Thus, the value of the integral is $\int_{0}^{1} \int_{(\sqrt{3} / 3) y}^{\sqrt{4 - y^{2}}} \sqrt{x^{2} + y^{2}} d x d y = \frac{4 π}{9}$

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Sketch the region of integration for each of integrals in Exercises 57–60, and then evaluate the integral by converting to polar coordinates.
$\int_{0}^{1} \int_{(\sqrt{3} / 3) y}^{\sqrt{4 - y^{2}}} \sqrt{x^{2} + y^{2}} d x d y$

Short Answer

Step by step solution

Given information

Calculation

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Sketch the region of integration for each of integrals in Exercises 57–60, and then evaluate the integral by converting to polar coordinates.∫01∫(3/3)y4-y2x2+y2dxdy

Short Answer

Step by step solution

Given information

Calculation

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Discrete Mathematics

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Mechanics Maths

Study anywhere. Anytime. Across all devices.

Sketch the region of integration for each of integrals in Exercises 57–60, and then evaluate the integral by converting to polar coordinates.
$\int_{0}^{1} \int_{(\sqrt{3} / 3) y}^{\sqrt{4 - y^{2}}} \sqrt{x^{2} + y^{2}} d x d y$