How do you find the area of the region cut from the second quadrant by the cardioid r=1-cosθ?

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Answer 1 · 2016-07-06T11:43:37

#= 3/4 pi + 2 #

that's

#A = int_(pi/2)^(pi) r^2 \ d theta#

#= int_(pi/2)^(pi) (1- cos theta)^2 \ d theta#

#= int_(pi/2)^(pi) d theta qquad 1 - 2 cos theta + cos^2 theta #

#= int_(pi/2)^(pi) d theta qquad 1 - 2 cos theta + (cos 2 theta +1)/2#

#= [ 3/2 theta - 2 sin theta + 1/4sin 2 theta ]_(pi/2)^(pi)#

#= [ 3/2 pi] - [ 3/4 pi - 2 ]#

#= 3/4 pi + 2 #