Chapter 13: Q. 63 (page 991)

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

Short Answer

Expert verified

The area of the circle is

$A = π R^{2}$ P

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 $r = 2 R \cos θ i s π R^{2} .$

calculation

Draw the circle

Plot of $r = 2 R \cos θ$

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.

$A = 2 \int_{6}^{a_{1}} \int_{5}^{x} r d r d θ$

Here, $θ_{1} = 0, θ_{2} = \frac{π}{2} a n d r_{1} = 0, r_{2} = r$

$A = 2 \int_{0}^{π / 2} \int_{0}^{r - 2 π \cos θ} r d r d θ$

Integrate with respect to r first

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

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

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

Integrate with respect to $θ$

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

localid="1650471847418" $A = π R^{2}$

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

Short Answer

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Use a double integral to prove that the area of the circle with radius R and equationr=2RcosθisπR2.

Short Answer

Step by step solution

Given information

calculation

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