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

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