Find the area between the circle r = 5 cos ? and the limacon r = 2 + cos ?. Also, please sketch the curves as well as the enclosed region.
The Correct Answer and Explanation is :
To determine the area enclosed between the circle ( r = 5 \cos \theta ) and the limaçon ( r = 2 + \cos \theta ), we first identify their points of intersection by equating the two equations:
[ 5 \cos \theta = 2 + \cos \theta ]
Simplifying, we get:
[ 4 \cos \theta = 2 ]
[ \cos \theta = \frac{1}{2} ]
This yields ( \theta = \pm \frac{\pi}{3} ) as the points of intersection.
The area ( A ) between two polar curves from ( \theta = a ) to ( \theta = b ) is given by:
[ A = \frac{1}{2} \int_a^b \left[ r_1^2(\theta) – r_2^2(\theta) \right] d\theta ]
Here, ( r_1(\theta) = 5 \cos \theta ) and ( r_2(\theta) = 2 + \cos \theta ).
Since the curves are symmetric about the polar axis, we can calculate the area in the interval ( [0, \frac{\pi}{3}] ) and then double it to account for the symmetric interval ( [-\frac{\pi}{3}, 0] ).
Thus, the total area is:
[ A = 2 \times \frac{1}{2} \int_0^{\frac{\pi}{3}} \left[ (5 \cos \theta)^2 – (2 + \cos \theta)^2 \right] d\theta ]
Simplifying:
[ A = \int_0^{\frac{\pi}{3}} \left[ 25 \cos^2 \theta – (4 + 4 \cos \theta + \cos^2 \theta) \right] d\theta ]
[ A = \int_0^{\frac{\pi}{3}} \left[ 24 \cos^2 \theta – 4 – 4 \cos \theta \right] d\theta ]
Using the double-angle identity ( \cos^2 \theta = \frac{1 + \cos 2\theta}{2} ):
[ A = \int_0^{\frac{\pi}{3}} \left[ 24 \left( \frac{1 + \cos 2\theta}{2} \right) – 4 – 4 \cos \theta \right] d\theta ]
[ A = \int_0^{\frac{\pi}{3}} \left[ 12 + 12 \cos 2\theta – 4 – 4 \cos \theta \right] d\theta ]
[ A = \int_0^{\frac{\pi}{3}} \left[ 8 + 12 \cos 2\theta – 4 \cos \theta \right] d\theta ]
Integrating term by term:
[ A = \left[ 8\theta + 6 \sin 2\theta – 4 \sin \theta \right]_0^{\frac{\pi}{3}} ]
Evaluating at the bounds:
[ A = \left( 8 \times \frac{\pi}{3} + 6 \sin \left( \frac{2\pi}{3} \right) – 4 \sin \left( \frac{\pi}{3} \right) \right) – (0 + 0 – 0) ]
[ A = \frac{8\pi}{3} + 6 \times \frac{\sqrt{3}}{2} – 4 \times \frac{\sqrt{3}}{2} ]
[ A = \frac{8\pi}{3} + 3\sqrt{3} – 2\sqrt{3} ]
[ A = \frac{8\pi}{3} + \sqrt{3} ]
Therefore, the area enclosed between the circle and the limaçon is ( \frac{8\pi}{3} + \sqrt{3} ) square units.
To visualize these curves, plot ( r = 5 \cos \theta ), which is a circle centered at ( (2.5, 0) ) with a radius of 2.5, and ( r = 2 + \cos \theta ), a limaçon with an inner loop. The enclosed region lies between these two curves.
For a visual explanation, you might find this video helpful: