Q. 64 Use a double integral to prove t... [FREE SOLUTION]

Chapter 13: Q. 64 (page 1028)

Use a double integral to prove that the area of the circle with radius $R$ and equation $r = 2 R \sin 胃$ is $蟺 R^{2}$ .

Short Answer

Expert verified

The area of the circle is $A = 蟺 R^{2}$

Step by step solution

Given information

The objective of this problem is to use double integral to prove that the area of the circle with radius $R$ and equation $r = 2 R \sin 胃$ is $蟺 R^{2}$ .

Calculation

Draw the circle

Plot of $r = 2 R \sin 胃$

Given circle is symmetrical about the horizontal axis. Therefore area of circle in polar form can be expressed as the twice of area of upper half circle.

$A = 2 {鈭�}_{a}^{a_{j}} {鈭�}_{n}^{n_{2}} r d r d 胃$

Here, $胃_{1} = 0, 胃_{2} = \frac{蟺}{2}$ and $r_{1} = 0, r_{2} = r$ $A = 2 {鈭�}_{0}^{蟺 / 2} {鈭�}_{0}^{r - 2 蟺 \sin 胃} r d r d 胃$

Integrate with respect to $r$ first

$A = 2 {鈭�}_{0}^{蟺 / 2} {[\frac{r^{2}}{2}]}_{0}^{2 R \sin 胃} d 胃 [鈭� x^{n} d x = \frac{x^{n + 1}}{n + 1} + C]$

$A = 2 {鈭�}_{0}^{x / 2} [\frac{(2 R \sin 胃)^{2} - 0}{2}]$

$A = 2 R^{2} {鈭�}_{0}^{蟺 / 2} [2 \sin^{2} 胃] d 胃 A = 2 R^{2} {鈭�}_{0}^{蟺 / 2} [1 - \cos 2 胃] d 胃$

Integrate with respect to $胃$

$A = 2 R^{2} {[胃 - \frac{1}{2} \sin 2 胃]}_{0}^{x / 2} [鈭� \cos x d x = \sin x + C] A = 2 R^{2} [\frac{蟺}{2} - \frac{1}{2} \sin 蟺 - 0] A = 蟺 R^{2}$

Thus, the area of the circle is

$A = 蟺 R^{2}$

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91影视

Use a double integral to prove that the area of the circle with radius $R$ and equation $r = 2 R \sin 胃$ is $蟺 R^{2}$ .

Short Answer

Step by step solution

Given information

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91影视

Use a double integral to prove that the area of the circle with radius Rand equation r=2Rsin胃is蟺R2.

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

Calculus

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Study anywhere. Anytime. Across all devices.

Use a double integral to prove that the area of the circle with radius $R$ and equation $r = 2 R \sin 胃$ is $蟺 R^{2}$ .