How to find the area of the loop of the curve #x(x^2+y^2)=a(x^2-y^2)#?

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Answer 1 · 2017-05-11T22:35:49

#(a^2(4-pi))/2#

Converting to polar using #x=rcostheta# and #y=rsintheta#:

#rcostheta(r^2cos^2theta+r^2sin^2theta)=a(r^2cos^2theta-r^2sin^2theta)#

#r^3costheta(cos^2theta+sin^2theta)=ar^2(cos^2theta-sin^2theta)#

Using #cos^2theta+sin^2theta=1#:

#r^3costheta=ar^2(cos^2theta-(1-cos^2theta))#

#r^3costheta-ar^2(2cos^2theta-1)=0#

#r^2(rcostheta-a(2cos^2theta-1))=0#

Note that #r^2=0# is just the point #(0,0)#, so we're left with:

#rcostheta=a(2cos^2theta-1)#

#r=a(2costheta-sectheta)#

The loop will begin and end when #r=0#, which is when

#2costheta=sectheta#

#2cos^2theta=1#

#costheta=pm1/sqrt2#

#theta=pmpi/4#

The area of a polar curve #r# from #theta=alpha# to #theta=beta# is given by #1/2int_alpha^betar^2d theta#, so here the integral for the area is:

#1/2int_(-pi/4)^(pi/4)[a(2costheta-sectheta)]^2d theta#

Since the loop is symmetric across the #x# axis:

#=int_0^(pi/4)a^2(2costheta-sectheta)^2d theta#

#=a^2int_0^(pi/4)(4cos^2theta-4costhetasectheta+sec^2theta)d theta#

Using #cos^2theta=1/2(1+cos2theta)#:

#=a^2int_0^(pi/4)(2(1+cos2theta)-4+sec^2theta)d theta#

#=a^2int_0^(pi/4)(2cos2theta+sec^2theta-2)d theta#

Integrating term by term:

#=a^2(sin2theta+tantheta-2theta)|_0^(pi/4)#

#=a^2(sin(pi/2)+tan(pi/4)-pi/2)-a^2(sin0+tan0-0)#

#=a^2(1+1-pi/2)#

#=(a^2(4-pi))/2#