Where #a=1#, the curve looks like:
Increasing or decreasing the value of #a# will only change the radius of the curve.
To find when the curve begins and ends, set #r=0#, since this is where the curve is at the origin.
If #asin3theta=0#, then #sin3theta=0#. Since #sintheta=0# at #theta=0,pi,2pi...# we see that for #sin3theta#, it will be #0# at #0,pi//3,2pi//3...#
So, the curve in the first quadrant varies from #theta=0# to #theta=pi//3#.
The expression for the area of any polar equation #r# from #theta=alpha# to #theta=beta# is given by #1/2int_alpha^betar^2d theta#.
For one loop of the given equation, the corresponding integral is then #1/2int_0^(pi//3)(asin3theta)^2d theta#.
Working this integral:
#1/2int_0^(pi//3)(asin3theta)^2d theta=1/2int_0^(pi//3)a^2(sin^2 3theta)d theta#
Use the identity #cos2alpha=1-2sin^2alpha# to rewrite for #sin^2alpha#, showing that #sin^2alpha=1/2(1-cos2alpha)#.
We can use this to say that #sin^2 3theta=1/2(1-cos6theta)#. Then the integral reduces:
#=1/2int_0^(pi//3)a^2(1/2(1-cos6theta))d theta=a^2/4int_0^(pi//3)(1-cos6theta)d theta#
Integrating term by term:
#=a^2/4(theta-1/6sin6theta)|_0^(pi//3)#
#=a^2/4[pi/3-1/6sin2pi-(0-1/6sin0)]#
#=a^2/4(pi/3)#
#=(pia^2)/12#
