x^2+y^2=ax =>(x-a/2)^2 + y^2 = (a/2)^2
是在x^2+y^2=a^2的内部
设x = r cost ,y = rsint代入x^2+y^2 = a^2得r=a
代入x^2+y^2=ax得 r^2 = arcost 所以r=acost
所以r的积分限为(acost,a)
∫∫√(x^2+y^2)dxdy
= ∫∫r^2drdt = ∫ 1/3a^3 - 1/3a^3cos^3t dt = 1/3a^3 * 1/6 (3π-4) = (3π-4)a^3 / 18