How do we find the area inside the cardioid $f(theta)=a(1+costheta)$ ?

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Answer 1 · 2016-05-27T08:27:17

Area of the region inside cardioid is $(3pia^2)/2$

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)

$int_0^(2pi)1/2(a(1+costheta))^2d theta$

= $a^2/2|theta+2sintheta+theta/2+1/4sin2theta|_0^(2pi)$

= $a^2/2(2pi+2sin(2pi)+(2pi)/2+1/4sin(4pi)-0-2sin0-0/2-1/4sin(4pi))$

= $a^2/2xx3pi=(3pia^2)/2$

How do we find the area inside the cardioid f(theta)=a(1+costheta)f(theta)=a(1+costheta)?