The Arc Length, L is given:

from parametric length equation we have the familiar:

#s = int_a^b sqrt(((dx)/dt)^2 + ((dy)/dt)^2) dt # it is just the aggregate distance formula that you may have seen many time. If you think the polar coordinate as a parametric representation in #r, alpha# then:

#r = f(alpha)#

#x= rcosalpha=f(alpha) cosalpha#

# y=rsinalpha=f(alpha) sinalpha #

#[(dx)/(dalpha)]^2 = [f'(alpha)cosalpha - f(alpha) sin(alpha)]^2 #

#[(dy)/(dalpha)]^2 = [f'(alpha)sinalpha + f(alpha) cos(alpha)]^2 #

Add and Simplify:

#[(dx)/(dalpha)]^2+[(dy)/(dalpha)]^2= f'^" 2"(alpha) + f^2(alpha) #

#L = int_(alpha1)^(alpha2) sqrt(f'^" 2"(alpha) + f^2(alpha)) " "dalpha #

Now evaluate:

# f'^" 2"(alpha) =[ d/(dalpha)(cos(3alpha-pi/2) + alphacsc(-alpha))]^2 #

# f^2(alpha) =[ cos(3alpha-pi/2) + alphacsc(-alpha)]^2 #

Integrand, #I =sqrt( f'^" 2"(alpha) + f^2(alpha)); alpha = t#

#I =sqrt(3cos3t + ((t cott - 1)csct)^2 + (sin3t - tcsct)^2 #

#L = int_(pi/12)^(pi/8)(I) dt #

At this point you ca use an Online Integration Calculator and get

#L ~~ .2037 #