For Polar Coordinates, the formula for the area A:
Given #r=theta-theta*sin ((7theta)/8)-cos((5theta)/3+pi/3)#
#A=1/2 int_alpha^beta r^2 *d theta#
#A=1/2 int_(pi/6)^((3pi)/2) (theta-theta*sin((7theta)/8)-cos((5theta)/3+pi/3))^2 d theta#
#A=1/2 int_(pi/6)^((3pi)/2) [theta^2+theta^2*sin^2((7theta)/8)+cos^2((5theta)/3+pi/3)#
#-2*theta^2*sin((7theta)/8)+2*theta*cos((5theta)/3+pi/3)*sin((7theta)/8)##-2*theta*cos((5theta)/3+pi/3)]d theta#
After some trigonometric transformation and integration by parts, it follows
#A=1/2[theta^3/3+theta^3/6-2/7*theta^2*sin((7theta)/4)-16/49*theta*cos((7theta)/4)+64/343*sin((7theta)/4)+theta/2+3/20*sin((10theta)/3+(2pi)/3)#
#+16/7*theta^2*cos((7theta)/8)-256/49*theta*sin((7theta)/8)-2048/343*cos((7theta)/8)-24/61*theta*cos((61theta)/24+pi/3)+576/3721*sin((61theta)/24+pi/3)#
#+24/19*theta*cos((19theta)/24+pi/3)-576/361*sin((19theta)/24+pi/3)##-6/5*theta*sin((5theta)/3+pi/3)-18/25*cos((5theta)/3+pi/3)]_(pi/6)^((3pi)/2)#
#A=1/2*[43.22026786-(-7.386403099)]#
#A=1/2*(50.60667096)#
#color (red)("Area A"=25.303335481" ""square units")#
God bless....I hope the explanation is useful.
