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