What is the area under the polar curve #f(theta) = thetasin((3theta)/4 )-cos^3((5theta)/12-pi/2) # over #[0,2pi]#?

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Answer 1 · 2016-11-02T17:16:31

I used WolframAlpha to evaluate:
#A = int_0^(2pi)(thetasin((3theta)/4) - cos^3((5theta)/12 - pi/2))^2d theta ~~ 45.592#

From the reference Area with polar coordinates , I obtained the equation:

#A = int_alpha^betar^2d theta#

where #r = f(theta)#

In this case, #r^2 = (thetasin((3theta)/4) - cos^3((5theta)/12 - pi/2))^2#

Substitute into the integral:

#A = int_0^(2pi)(thetasin((3theta)/4) - cos^3((5theta)/12 - pi/2))^2d theta#