How do you find the area of the region shared by the circles #r=2cos(theta)# and #r=2sin(theta)#?

1 Answer
Oct 24, 2015

Answer:

#pi/2-1#

Explanation:

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Intersection:
#2sintheta=2costheta => sintheta/costheta=1 => tantheta=1 => theta=pi/4#

#A=intint_A rdrd theta#

#A=int_0^(pi/4) d theta int_0^(2sintheta) rdr + int_(pi/4)^(pi/2) d theta int_0^(2costheta) rdr#

#A_1=1/2 int_0^(pi/4) d theta * 4sin^2 theta = int_0^(pi/4) (1-cos2theta) d theta#

#A_1 = (theta-1/2sin2theta) |_0^(pi/4) = pi/4-1/2#

#A_2=1/2 int_(pi/4)^(pi/2) d theta * 4cos^2 theta = int_(pi/4)^(pi/2) (1+cos2theta) d theta#

#A_2 = (theta+1/2sin2theta) |_(pi/4)^(pi/2) = pi/4-1/2#

#A=pi/4-1/2+pi/4-1/2 = pi/2-1#