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

1 Answer
May 27, 2016

Answer:

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

Explanation:

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#