How do you find the area of one petal of #r=cos2theta#?

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Answer 1 · 2016-11-30T13:00:38

#pi/8=0.3927# areal units, nearly.

Period of #r(theta)# is #(2pi)/2=pi#.

As # r = cos 2theta >= 0, 2theta in [-pi/2, pi/2] to theta in [-pi/4, pi/4]#,

for one petal. So, the area (by symmetry about #theta = 0)

#=2(1/2 int r^2 d theta)#, from #0 to pi/4#

#=int cos^ 2 2theta d theta #, for the limits

#=int (1+cos 4theta)/2 d theta #, for the limits

=#[1/2[theta+1/4sin 4theta]#, between #theta = 0 and theta = pi/4#

#=pi/8# areal units.

graph{(x^2+y^2)sqrt(x^2+y^2)=x^2-y^2 [-2.5, 2.5, -1.25, 1.25]}