What is the area under the polar curve #f(theta) = theta^2-thetasin(2theta-pi/4 ) +cos(3theta-(5pi)/4)# over #[pi/8,pi/2]#?

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Answer 1 · 2016-10-14T17:22:23

#A ~~ 0.65#

#A = int_(alpha)^(beta)(1/2)r^2d theta#

Substitute the #f(theta)# for r:

#A = (1/2)int_(pi/8)^(pi/2)(theta^2 - thetasin(2theta- pi/4) + cos(3theta - (5pi)/4))^2d theta#

#A ~~ 0.65#