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Find the area of the region that lies inside the first curve and outside the second curve.

This question aims to find the area of the region that lies inside the first curve and outside the second curve.

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)$.

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.

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