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Higher mathematics surface integral?
The surface integral of coordinates does not have the properties you said. This question applies Gaussian formula and obtains.

Original formula = ∫∫∫; (0+y-z)dxdydz =∫& lt; 0,3 & gtdz∫& lt; 0,2π& gt; dt∫& lt; 0, 1 & gt; (rsint-z)rdr

=∫& lt; 0,3 & gtdz∫& lt; 0,2π& gt; dt[( 1/3)r^3sint-( 1/2)zr^2]<; 0, 1 & gt;

=∫& lt; 0,3 & gtdz∫& lt; 0,2π& gt; [( 1/3)Sint-( 1/2)z]dt

=∫& lt; 0,3 & gtdz[-( 1/3)cost-( 1/2)ZT]& lt; 0,2π& gt;

=∫& lt; 0,3 & gt(-πz)dz =(-π/2)[z^2]<; 0,3 & gt= -9π/2

Not projection calculation, but degaussing formula:

& lt∑& gt; pdy dz+qd zdx+RDX dy =∫∫∫& lt; ω& gt; (? P/? x+? Q/? y+? R/? z)dxdydz