This question aims to find the area of the region that lies inside the first curve and outside the second curve.
Circle
The area of the region can be found by subtraction. We can subtract the area of the first circle from the second circle. For polar curves, we can get the area from the radius $r= f (\theta)$ and $ r = g (\theta)$.
Radius of circle
Subtraction
There are two curves with two different radiuses. These are as follows:
\[ R = 7 \]
\[ R = 14 cos \theta \]
Expert Answer
By equating both radiuses:
\[ 14 cos \theta = 7 \]
\[ cos \theta = \frac { 7 } { 14 } \]
\[ cos \theta = \frac { 1 } { 2 } \]
\[ \theta = cos ^{-1}\frac { 1 }{ 2 } \]
\[ \theta = \frac { \pi } { 3 } \]
The limits are 0 and $ \frac { \pi } { 3 } $
The area of the region can be calculated by:
\[ A = \int_{ 0 }^{ \frac { \pi } { 3 } } ( 14 cos \theta ) ^ 2 – 7 ^ 2 \, d\theta \]
\[ A = \int_{ 0 }^{ \frac { \pi } { 3 } } ( 196 cos ^ 2 \theta – 49) \, d\theta \]
\[ A = 196 \int_{ 0 }^{ \frac { \pi } { 3 } } cos ^ 2 \theta \, d\theta – 49 \int_{ 0 }^{ \frac { \pi } { 3 } } r \, d\theta \]
\[ A = [ 98 \theta + 98 sin ( 2 \theta ) ] _ 0 ^ { \frac {\pi}{3} } – 49 [ \theta ] _ 0 ^ { \frac {\pi}{3} } \]
\[ A = [ 98 ( \frac {\pi}{3} – 0 ) + 98 sin ( 2 (\frac {\pi}{3})) – 49 sin ( 2 ( 0 ) ) ] – 49 [\frac {\pi}{3}] – 0 \]
\[ A = [ 49 ( \frac { \sqrt { 3 }} { 2 } – 49 ( 0 ) ] + 49 [ \frac { \pi } { 3 } ] \]
\[ A = \frac { 49 \sqrt 3 } { 2 } + \frac { 49 \pi } { 3 } \]
\[ A = 93, 7479 \]
Numerical Solution
The area of the region that lies inside the first curve and outside the second curve is 93, 7479.
Example
Calculate the area inside and outside the unit circle having function $ f (\theta) = 2 cos ( \theta ) $ and $ g ( \theta ) = 1 $
\[ cos \theta = \frac { 1 } { 2 } \]
\[ \theta = cos ^ {-1} \frac { 1 } { 2 } \]
\[ \theta = \pm \frac { \pi } { 3 } \]
The limits are $ – \frac { \pi } { 3 } $ and $ \frac { \pi } { 3 } $
The area of the region can be calculated by:
\[ A = \frac { 1 } { 2 } \int_{ – \frac { \pi } { 3 } } ^ { \frac { \pi } { 3 } } [ ( 2 cos ( \theta) ) ^ 2 – 1 ^ 2 ] d \theta \]
\[A = \frac { 1 } { 2 } ( \theta + sin 2 ( \theta ) )| _ {-\frac { \pi}{3}} ^ {\frac { \pi}{3}} \]
\[ A = \frac { \pi } { 3 } + \frac { \sqrt {3}}{2} \]
\[ A = 1.91\]
Image/Mathematical drawings are created in Geogebra.