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How to find dz in mathematics
The whole domain ω is an inverted tetrahedron with the origin as the lower vertex,

The other three vertices are (1, 0, 1), (0, 1, 1), (0,0, 1),

Exchange the order of integrals and get

I =∫& lt; 0, 1 & gt; (sinz/z)dz∫& lt; 0,z & gtdy∫& lt; 0,z-y & gt; Advanced (short for deluxe)

=∫& lt; 0, 1 & gt; (sinz/z)dz∫& lt; 0,z & gt(z-y)dy

=∫& lt; 0, 1 & gt; (sinz/z)dz[zy-y^2/2]<; y=0,y = z & gt

=∫& lt; 0, 1 & gt; (sinz/z)(z^2/2)dz =( 1/2)∫& lt; 0, 1 & gt; Zinczz

=(- 1/2)∫& lt; 0, 1 & gt; zdcosz

=(- 1/2){[zcosz]& lt; 0, 1 & gt; +[sinz]& lt; 0, 1 & gt; }

=( 1/2)(sin 1-cos 1)