 # Use a double integral to find the area of the region inside the circle and outside the circle.

Region inside the circle is represented by $(x-5)^{2}+y^{2}=25$

Region outside the circle $x^{2}+y^{2}=25$

This question aims to find the area under the region of the circle. The area of a region inside or outside the circle can be found by using a double integral and integrating the function over the region. Polar coordinates are sometimes easy to integrate as they simplify the limits of integration.

Step 1

A basic understanding of equations tells us that this equation is a circle shifted five units to the right.

$(x-5) ^ {2} + y ^ {2} = 25$

$(r \cos \theta – 5) ^ {2} + (r^{2} \sin ^ {2} \theta)=25$

$( r^ {2} \ cos ^{2} \theta – 10r \cos \theta + 25)+(r ^{2} \sin^{2} \theta) = 25$

$r^ {2}. \cos ^{2} \theta + r^{2} \sin ^{2}. \theta = 10.r \cos \theta$

$x ^{2} +y ^ {2} = 10r \cos \theta$

$r ^{2} = 10r \cos \theta$

$r = 10 \cos \theta$

Step 2

Again, understanding that this is the equation of a circle with a radius of $5$ is helpful.

$x ^{2} + y ^{2} = 25$

$r ^{2} = 25$

$r = 5$

Step 3

Determine the limits of integration:

$5 = 10 \cos \theta$

$\cos \theta = \dfrac{5}{10}$

$\cos \theta = \dfrac{1}{2}$

$\theta = (0, \dfrac {\pi} {3}) , (0, \dfrac{\pi}{3})$

Step 4

Our region can be defined as:

$R = (r, \theta) | (0,\dfrac {\pi} {3} ) ,(0, \dfrac {\pi} {3})$

Step 5

Set up the integral:

$Area=2 \int _{0} ^ {\dfrac {\pi}{3}} \dfrac {1}{2} (10 \cos \theta )^{2} d\theta – 2\int_{0} ^{\dfrac {\pi} {3}} (\dfrac {1}{2}) (5)^{2} d\theta$

Step 6

Integrate with respect to:

$=\int _{0} ^ {\dfrac {\pi}{3}} (100 \cos \theta )d\theta – \int_{0} ^{\dfrac {\pi} {3}} 25 d\theta$

Step 7

$=50 ( \theta + \dfrac {sin2\theta}{2})|_{0} ^{\dfrac{\pi}{3}} -(25) |_{0}^{\dfrac {\pi}{3}}$

$=50(\dfrac{\pi}{3} + \dfrac {1}{2}.\dfrac{\sqrt 3}{2}) – (\dfrac{25\pi}{3})$

Step 8

$Area=\dfrac{25\pi}{3} + \dfrac{25 \sqrt 3}{2}$

## Numerical Result

The area of the region is $\dfrac{25\pi}{3} + \dfrac{25 \sqrt 3}{2}$.

## Example

Use double integral to determine the area of the region. The region inside the circle $(x−1)^{2}+y^{2}=1$ and outside the circle $x^{2} +y^{2}=1$.

Solution

Step 1

$(x-1) ^ {2} + y ^ {2} = 1$

$(r \cos \theta – 1) ^ {2} + (r^{2} \sin ^ {2} \theta)=1$

$( r^ {2} \ cos ^{2} \theta – 2r \cos \theta + 1)+(r ^{2} \sin^{2} \theta) = 1$

$r^ {2}. \cos ^{2} \theta+ r^{2}. \sin ^{2} \theta=2r \cos \theta$

$x ^{2} +y ^ {2} = 2r \cos \theta$

$r ^{2} = 2r \cos \theta$

$r = 2\cos \theta$

Step 2

$x ^{2} + y ^{2} = 1$

$r ^{2} = 1$

$r = 1$

Step 3

Determine the limits of integration:

$1= 2\cos \theta$

$\cos \theta = \dfrac{1}{2}$

$\cos \theta = \dfrac{1}{2}$

$\theta = (0, \dfrac {\pi} {3}) , (0, \dfrac{\pi}{3})$

Step 4

Our region can be defined as:

$R = (r, \theta) | (0,\dfrac {\pi} {3} ) ,(0, \dfrac {\pi} {3})$

Step 4

Integrate the region and plug the limits of integration result in the area of the region.

$Area=\dfrac{\pi}{3} + \dfrac{\sqrt 3}{2}$

5/5 - (17 votes)