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Find the area of the region that is inside r=3cos(Θ) and outside r=2-cos(Θ).

This article aims to find the area under the given curves. The article uses the background concept of the area under the curve and integration. The area under curve can be calculated in three simple steps. First, we need to know equation of the curve $(y = f(x))$, the limits over which area is to be calculated, and axis bounding the area. Second, we need to find the integration (antiderivative) of the curve. Finally, we need to apply an upper and lower bound to the integral response and take the difference to get the area under the curve.

Expert Answer

\[r = 3 \cos\theta\]

\[r = 2-\cos\theta\]

First, find the intersections.

\[3\cos\theta = 2-\cos\theta\]

\[4 \cos\theta = 2\]

\[\cos\theta = \dfrac{1}{2}\]

\[\theta = \dfrac{-\pi}{3} , \dfrac{\pi}{3}\]

We want the area inside the first curve and outside the second curve. So $R = 3 \cos\theta $ and $r = 2 – \cos\theta $, so $R > r$.

Now integrate to find the final answer.

\[A = \int \dfrac{1}{2} (R^{2} – r^{2})d\theta \]

\[A = \int \dfrac{1}{2} ((3\cos\theta)^{2} – (2-\cos\theta)^{2})d\theta \]

\[A = \int \dfrac{1}{2} ((9\cos^{2}\theta) – (4-4\cos\theta+cos^{2}\theta))d\theta \]

\[A= \int \dfrac{1}{2} (8\cos^{2}\theta +4\cos\theta-4) d\theta\]

\[A= \int (4\cos^{2}\theta +2\cos\theta-2) d\theta\]

Using power reduction formula.

\[A = \int (2+2\cos(2\theta)+2\cos\theta -2) d\theta\]

\[A = \int (2\cos(2\theta)+2\cos\theta)d\theta\]

Integrating

\[A = |\sin(2\theta) + 2\sin\theta |_{\dfrac{-\pi}{3}}^{\dfrac{\pi}{3}}\]

\[A = 3\sqrt 3\]

The area inside of $ r = 3\cos\theta $ and outside of $ r = 2-\cos\theta$ is $3\sqrt 3$.

Numerical Result

The area inside of $ r = 3\cos\theta $ and outside of $ r = 2-\cos\theta$ is $3\sqrt 3$.

Example

Find the area of ​​region that is inside $r=5\cos(\theta)$ and outside $r=2+\cos(\theta)$.

Example

\[r = 5 \cos\theta\]

\[r = 2 + \cos \theta\]

First, find the intersections.

\[5\cos\theta = 2+\cos\theta\]

\[4 \cos\theta = 2\]

\[\cos\theta = \dfrac{1}{2}\]

\[\theta = \dfrac{-\pi}{3} , \dfrac{\pi}{3}\]

We want the area inside the first curve and outside the second curve. So $ R  =  5 \cos \theta $ and $ r  =  2 + \cos\theta $, so $ R  >  r $.

Now integrate to find the final answer.

\[A = \int \dfrac{1}{2} (R^{2} – r^{2})d\theta \]

\[A = \int \dfrac{1}{2} ((5\cos\theta)^{2} – (2+\cos\theta)^{2})d\theta \]

\[A = \int \dfrac{1}{2} ((25\cos^{2}\theta) – (4+4\cos\theta+cos^{2}\theta))d\theta \]

\[ A = \int \dfrac{ 1 } { 2 } ( 25 \cos ^ { 2 } \theta – 4 – 4 \cos \theta – cos ^ { 2 } \theta ) ) d \theta \]

\[ A = \int \dfrac{ 1 } { 2 } ( 24 \cos ^ { 2 } \theta – 4 \cos \theta – 4 ) d\theta \]

\[ A = \int ( 12 \cos ^ { 2 } \theta \: – \: 2 \cos \theta \: -\: 2 ) d \theta\]

Using power reduction formula.

\[ A = \int ( 6 + 6 \cos ( 2 \theta ) – 2 \cos \theta – 2 ) d \theta\]

\[ A = \int ( 4 + 6 \cos( 2 \theta ) – 2 \cos \theta ) d \theta\]

Integrating

\[A = |4\theta +3 \sin ( 2\theta) – 2\sin\theta |_{\dfrac{-\pi}{3}}^{\dfrac{\pi}{3}}\]

\[A = \dfrac{8\pi}{3}-\sqrt 3\]

The area inside of $ r = 5 \cos \theta $ and outside of $ r  =   2 + \cos \theta $ is $ \dfrac{8\pi}{3}-\sqrt 3 $.

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