Question

求二重积分∫∫(x^2-y^2)dxdy,D为0≤y≤sinx,0≤x≤π所围成的区域,需画图

Answer 1

∵∫x²sinxdx=(-x²cosx+2xsinx+2cosx)│ (应用分部积分法)

=π²-2-2

=π²-4

∫sin³xdx=∫(1-cos²x)sinxdx

=∫(cos²x-1)d(cosx)

=(cos³x/3-cosx)│

=-1/3+1-1/3+1

=4/3

∴∫∫(x²-y²)dxdy=∫dx∫(x²-y²)dy

=∫(x²sinx-sin³x/3)dx

=∫x²sinxdx-(1/3)∫sin³xdx

=(π²-4)-(1/3)(4/3)

=π²-40/9.