dS=\1+x+zy dxdy =√1+0+(-1)2c=√2d, 故[(x+y+z)s 2(x+y+5-p)dd=2(5+x) v2 def (5+r cos O) rdr =125V2T.  故 (x + y + z)ds  = + + − Dxy 2 (x y 5 y)dxdy =  + Dxy 2 (5 x)dxdy d r rdr   =  +   5 0 2 0 2 (5 cos ) = 125 2. dS z z dxdy x y 2 2 = 1+  +  dxdy 2 = 1+ 0 + (−1) = 2dxdy