Area inside the cardioid or any other bound curve is found using integral
#int_0^(2pi)1/2(f(theta))^2d theta#
As here #f(theta)=a(1+costheta)#
Area is #int_0^(2pi)1/2(a(1+costheta))^2d theta#
= #a^2/2int_0^(2pi)(1+costheta)^2d theta#
= #a^2/2int_0^(2pi)(1+2costheta+cos^2theta)d theta#
= #a^2/2|theta+2sintheta|_0^(2pi)+a^2/2int_0^(2pi)cos^2thetad theta# (A)
Now #cos^2theta=1/2(1+cos2theta)#
Hence #int_0^(2pi)cos^2thetad theta=int_0^(2pi)(1/2+1/2cos2theta)d theta#
= #|theta/2+1/4sin2theta|_0^(2pi)# (B)
Hence, substituting (B) in (A)
#int_0^(2pi)1/2(a(1+costheta))^2d theta#
= #a^2/2|theta+2sintheta+theta/2+1/4sin2theta|_0^(2pi)#
= #a^2/2(2pi+2sin(2pi)+(2pi)/2+1/4sin(4pi)-0-2sin0-0/2-1/4sin(4pi))#
= #a^2/2xx3pi=(3pia^2)/2#
