Question

高等数学，二重积分

Answer 1

∣x+y∣的值关于x=-y对称，
原式=2∫【-√2/2,√2/2】dx∫【-x，√(1-x²)】(x+y)dy
= 2∫【-√2/2,√2/2】dx(xy+y²/2)【-x，√(1-x²)】
= 2∫【-√2/2,√2/2】dx( x√(1-x²)+x²+(1-x²)/2-x²/2)
= 2∫【-√2/2,√2/2】dx(x√(1-x²)+1/2)
= 2∫【-√2/2,√2/2】(x√(1-x²)dx+√2
设x=sin t,则t=arcsin x，
原式= 2∫【-π/4,π/4】sin t cos tdsint +√2
=∫【-π/4,π/4 】sin2tcostdt+√2
= ∫【-π/4,π/4】[(sin3 t+sint)/2]dt+√2
= (-cos3t/6-cost/2)【-π/4,π/4】 +√2
=√2

追答

∣x+y∣的值关于x=-y对称，
原式=2∫【-√2/2,√2/2】dx∫【-x，√(1-x²)】(x+y)dy +2∫【√2/2,1】xdx ∫【-√(1-x²),√(1-x²)】ydy
= 2∫【-√2/2,√2/2】dx(xy+y²/2)【-x，√(1-x²)】+2∫【√2/2,1】2x√(1-x²)dx
= 2∫【-√2/2,√2/2】dx( x√(1-x²)+x²+(1-x²)/2-x²/2) +2∫【√2/2,1】2x√(1-x²)dx
= 2∫【-√2/2,√2/2】dx(x√(1-x²)+1/2) +2∫【√2/2,1】2x√(1-x²)dx
= 2∫【-√2/2,√2/2】(x√(1-x²)dx+√2 +2∫【√2/2,1】2x√(1-x²)dx
设x=sin t,则t=arcsin x，
原式= 2∫【-π/4,π/4】sin t cos tdsint +√2+2∫【π/4,π/2】2sintcostdsint
=∫【-π/4,π/4 】sin2tcostdt+√2
= ∫【-π/4,π/4】[(sin3 t+sint)/2]dt+√2+2∫【π/4,π/2】sin2tcostdt
= (-cos3t/6-cost/2)【-π/4,π/4】 +√2+(-cos3t/3-cost)【π/4,π/2】
=√2+(-√2/6+√2/2)
=4√2/3

Answer 2