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.<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To determine the area enclosed between the circle ( r = 5 \\cos \\theta ) and the lima\u00e7on ( r = 2 + \\cos \\theta ), we first identify their points of intersection by equating the two equations:<\/p>\n\n\n\n [ 5 \\cos \\theta = 2 + \\cos \\theta ]<\/p>\n\n\n\n Simplifying, we get:<\/p>\n\n\n\n [ 4 \\cos \\theta = 2 ]<\/p>\n\n\n\n [ \\cos \\theta = \\frac{1}{2} ]<\/p>\n\n\n\n This yields ( \\theta = \\pm \\frac{\\pi}{3} ) as the points of intersection.<\/p>\n\n\n\n The area ( A ) between two polar curves from ( \\theta = a ) to ( \\theta = b ) is given by:<\/p>\n\n\n\n [ A = \\frac{1}{2} \\int_a^b \\left[ r_1^2(\\theta) – r_2^2(\\theta) \\right] d\\theta ]<\/p>\n\n\n\n Here, ( r_1(\\theta) = 5 \\cos \\theta ) and ( r_2(\\theta) = 2 + \\cos \\theta ).<\/p>\n\n\n\n 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] ).<\/p>\n\n\n\n Thus, the total area is:<\/p>\n\n\n\n [ A = 2 \\times \\frac{1}{2} \\int_0^{\\frac{\\pi}{3}} \\left[ (5 \\cos \\theta)^2 – (2 + \\cos \\theta)^2 \\right] d\\theta ]<\/p>\n\n\n\n Simplifying:<\/p>\n\n\n\n [ A = \\int_0^{\\frac{\\pi}{3}} \\left[ 25 \\cos^2 \\theta – (4 + 4 \\cos \\theta + \\cos^2 \\theta) \\right] d\\theta ]<\/p>\n\n\n\n [ A = \\int_0^{\\frac{\\pi}{3}} \\left[ 24 \\cos^2 \\theta – 4 – 4 \\cos \\theta \\right] d\\theta ]<\/p>\n\n\n\n Using the double-angle identity ( \\cos^2 \\theta = \\frac{1 + \\cos 2\\theta}{2} ):<\/p>\n\n\n\n [ A = \\int_0^{\\frac{\\pi}{3}} \\left[ 24 \\left( \\frac{1 + \\cos 2\\theta}{2} \\right) – 4 – 4 \\cos \\theta \\right] d\\theta ]<\/p>\n\n\n\n [ A = \\int_0^{\\frac{\\pi}{3}} \\left[ 12 + 12 \\cos 2\\theta – 4 – 4 \\cos \\theta \\right] d\\theta ]<\/p>\n\n\n\n [ A = \\int_0^{\\frac{\\pi}{3}} \\left[ 8 + 12 \\cos 2\\theta – 4 \\cos \\theta \\right] d\\theta ]<\/p>\n\n\n\n Integrating term by term:<\/p>\n\n\n\n [ A = \\left[ 8\\theta + 6 \\sin 2\\theta – 4 \\sin \\theta \\right]_0^{\\frac{\\pi}{3}} ]<\/p>\n\n\n\n Evaluating at the bounds:<\/p>\n\n\n\n [ 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) ]<\/p>\n\n\n\n [ A = \\frac{8\\pi}{3} + 6 \\times \\frac{\\sqrt{3}}{2} – 4 \\times \\frac{\\sqrt{3}}{2} ]<\/p>\n\n\n\n [ A = \\frac{8\\pi}{3} + 3\\sqrt{3} – 2\\sqrt{3} ]<\/p>\n\n\n\n [ A = \\frac{8\\pi}{3} + \\sqrt{3} ]<\/p>\n\n\n\n Therefore, the area enclosed between the circle and the lima\u00e7on is ( \\frac{8\\pi}{3} + \\sqrt{3} ) square units.<\/p>\n\n\n\n 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 lima\u00e7on with an inner loop. The enclosed region lies between these two curves.<\/p>\n\n\n\n For a visual explanation, you might find this video helpful:<\/p>\n\n\n\n