How do you find region bounded by circle r = 3cosΘ and the cardioid r = 1 + cosΘ?

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Answer 1 · 2015-10-25T08:28:36

#A= (5pi)/4#

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Lets find the intersection of the curves in the first quadrant:

#3costheta=1+costheta => 2costheta=1 => costheta=1/2 => theta=pi/3#

The region is symmetric so we can find the area of the half of it:

#A= 2 (int_0^(pi/3) d theta int_0^(1+costheta) rdr + int_(pi/3)^(pi/2) d theta int_0^(3costheta) rdr)#

#A_1= 1/2int_0^(pi/3) d theta r^2 |_0^(1+costheta) = 1/2int_0^(pi/3) d theta (1+2costheta+cos^2theta)#

#A_1= 1/2 int_0^(pi/3) d theta (1+2costheta+(1+cos2theta)/2)#

#A_1= 1/2 ((3theta)/2+2sintheta+1/4sin2theta)|_0^(pi/3) =pi/4+(9sqrt3)/16#

#A_2= 1/2int_(pi/3)^(pi/2) d theta r^2 |_0^(3costheta)=#

#1/2int_(pi/3)^(pi/2) d theta (9cos^2theta) = 9/4 int_(pi/3)^(pi/2) d theta (1+cos2theta) =#

#= 9/4 (theta +1/2sin2theta) |_(pi/3)^(pi/2) = (3pi)/8-(9sqrt3)/16#

#A=2(pi/4+(9sqrt3)/16 + (3pi)/8-(9sqrt3)/16) = 2(5pi)/8 = (5pi)/4#