设区域D为曲线y=x^2,y=1所围成.计算∫∫√y^2-x^2（根号完） dxdy,其中积分与为D

∫∫√(y-x²)dxdy=∫dx∫√(y-x²)dy
=∫{[(2/3)(y-x²)^(3/2)]│}dx
=(2/3)∫(1-x²)^(3/2)dx
=(2/3)∫(cos²t)²dt (令x=sint)
=(2/3)∫[(1+cos(2t))/2]²dt
=(1/6)∫[1+2cos(2t)+cos²(2t)]dt
=(1/6)∫[3/2+2cos(2t)+cos(4t)/2]dt
=(1/6)[3t/2+sin(2t)+sin(4t)/8]│
=(1/6)[(3/2)(π/2+π/2)]
=π/4.