问题问得很模糊,下面θ在[0,2*pi]内来计算：
令 x = y ==>θ1=pi/4,θ2=5*pi/4;
==>[pi/4,5*pi/4]内的面积
s(t) = [-1/3*sint^2*cost-2/3*cost-1/3*cost^2*sint-2/3*sint]*|a|
==>S1 = s(5*pi/4) - s(pi/4)
==>S1 = [1.1785+1.1785]|a| = 2.357|a|
==>在一个周期内 x 与 y 围成的面积
S = 2*S1 = 4.714|a|