Question

三重积分计算∫∫∫ycos(x+z)dv，Ω由y=√x,y=0,z=0,x+z=π/2围成的闭区

三重积分计算∫∫∫ycos(x+z)dv，Ω由y=√x,y=0,z=0,x+z=π/2围成的闭区域，求大神帮忙～

Answer 1

解：原式=∫<0,π/2>dx∫<0,√x>ydy∫<0,π/2-x>cos(x+z)dz
=∫<0,π/2>dx∫<0,√x>y(1-sinx)dy
=(1/2)∫<0,π/2>x(1-sinx)dx
=(1/2)[∫<0,π/2>xdx-∫<0,π/2>xsinxdx]
=(1/2)[π²/8-∫<0,π/2>xsinxdx]
=(1/2)(π²/8-1) (应用分部积分法)
=(π²-8)/16。

Answer 2

=∫<0,π/2>dx∫<0,√x>ydy∫<0,π/2-x>cos(x+z)dz
=∫<0,π/2>dx∫<0,√x>y(1-sinx)dy
=(1/2)∫<0,π/2>x(1-sinx)dx
=(1/2)[∫<0,π/2>xdx-∫<0,π/2>xsinxdx]
=(1/2)[π²/8-∫<0,π/2>xsinxdx]