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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.

Expert Answer

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.

\[Area=\dfrac{\pi}{3} + \dfrac{\sqrt 3}{2}\]

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