Question

计算二重积分I=∫∫√(y-x²)dxdy,其中D={（x,y）| |x|≤1，x²≤y≤1}

Answer 1

解：原式=∫<-1,1>dx∫<x²,1>√(y-x²)dy
=(2/3)∫<-1,1>(1-x²)^(3/2)dx
=(2/3)∫<-π/2,π/2>(cost)^4dt (令x=sint)
=(2/3)∫<-π/2,π/2>[3/8+cos(2t)/2+cos(4t)/8]dt
=(2/3)[3t/8+sin(2t)/4+cos(4t)/32]│<-π/2,π/2>
=(2/3)(3π/16+3π/16)
=π/4。

Answer 2

∫(-1,1)[∫(x^2,1)(y-x^2)dy]dx
=∫(-1,1)[(1/2-x^4/2)-x^2(1-x^2)]dx
=∫(-1,1)(1/2-x^2+x^4/2)dx
=1/2[1-(-1)]-1/3[1^3-(-1^)3)]+1/10[1^5-(-1)^5]
=1-2/3+1/5=8/15