Chapter 13: Q. 64 (page 1028)

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

Short Answer

Expert verified

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

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 $r = 2 R \sin θ$ is $π R^{2}$ .

Calculation

Draw the circle

Plot of $r = 2 R \sin θ$

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.

$A = 2 \int_{a}^{a_{j}} \int_{n}^{n_{2}} r d r d θ$

Here, $θ_{1} = 0, θ_{2} = \frac{π}{2}$ and $r_{1} = 0, r_{2} = r$ $A = 2 \int_{0}^{π / 2} \int_{0}^{r - 2 π \sin θ} r d r d θ$

Integrate with respect to $r$ first

$A = 2 \int_{0}^{π / 2} {[\frac{r^{2}}{2}]}_{0}^{2 R \sin θ} d θ [\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C]$

$A = 2 \int_{0}^{x / 2} [\frac{(2 R \sin θ)^{2} - 0}{2}]$

$A = 2 R^{2} \int_{0}^{π / 2} [2 \sin^{2} θ] d θ A = 2 R^{2} \int_{0}^{π / 2} [1 - \cos 2 θ] d θ$

Integrate with respect to $θ$

$A = 2 R^{2} {[θ - \frac{1}{2} \sin 2 θ]}_{0}^{x / 2} [\int \cos x d x = \sin x + C] A = 2 R^{2} [\frac{π}{2} - \frac{1}{2} \sin π - 0] A = π R^{2}$

Thus, the area of the circle is

$A = π R^{2}$

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

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Use a double integral to prove that the area of the circle with radius Rand equation r=2RsinθisπR2.

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