Q. 58 Sketch the region of integration... [FREE SOLUTION]

Chapter 13: Q. 58 (page 1028)

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

Short Answer

Expert verified

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

Step by step solution

Given information

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

Calculation

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

The foundation is

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

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 $x$

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

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

$\tan 胃 = \frac{1}{\sqrt{3}} 鈬� 胃 = \frac{蟺}{6}$

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

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

$r \sin 胃 = 0 鈬� r = 0, 胃 = 0$

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

$d x d y = r d r d 胃$

Therefore,

Integrate with respect to $x$ first

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

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91影视

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

Short Answer

Step by step solution

Given information

Calculation

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91影视

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

Short Answer

Step by step solution

Given information

Calculation

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Decision Maths

Geometry

Theoretical and Mathematical Physics

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Calculus

Logic and Functions

Study anywhere. Anytime. Across all devices.

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