How do you find the region inside cardioid #r=(alpha)(1+sin(theta)), (alpha)>0#?

1 Answer
Oct 16, 2015

Answer:

#A=(3pi alpha^2)/2#

Explanation:

#A=intint rdrd theta#

#A=int_0^(2pi) d theta int_0^(alpha(1+sin theta)) rdr#

#A=int_0^(2pi) d theta r^2/2|_0^(alpha(1+sin theta)) = 1/2 int_0^(2pi) (alpha (1+sin theta))^2 d theta#

#A = alpha^2/2 int_0^(2pi) (1+2sintheta+sin^2theta) d theta#

#A = alpha^2/2 int_0^(2pi) (1+2sintheta+sin^2theta) d theta#

#A = alpha^2/2 (theta-2costheta+1/2int_0^(2pi) (1-cos2theta) d theta) #

#A = alpha^2/2 (theta-2costheta+theta/2 - 1/4sin2theta) |_0^(2pi) => #

#A=alpha^2/2 (3pi-2+2)#

#A=(3pi alpha^2)/2#