=5F=于x+x2d (二=0,d== IG4cos0 sin 0+cos 02 cos0)de =1609=1641c9=1643,z=12x 解法2用高斯公式。补一块面S1:x2+y2≤4,下侧 noF:ns=∫nonF:ds-』mF:△ f3divrotFdrdyds:-[rotF nds (divroIF=O) Co ap rOtF·ndS 5, ax ay (rOF.n是F在方方向的投影,即rotF在z轴方向的分量乘以(-1) 3x2dS=「3x2ddhy=d「3r3cos2bbh +1-≤48  =  L F ds    = + L xdx x dy 3 (z = 0,dz = 0)       ( 4cos sin 8cos 2cos )d 2 0 3  = − +    d  = 2 0 4 16cos  =  2 0 4 16 4 cos  d   12 4 2 2 3 16 4 =  =   解法 2 用高斯公式。补一块面 1 S ： 4 2 2 x + y  ，下侧   S rotF ndS    + =  S S1 rotF dS    −  S1 rotF dS    = V divrotFdxdydz   −  S1 rotF ndS   (divrotF = 0)   = −  S1 rotF ndS      −   = 1 [ ] S dS y P x Q ( rotF n    是 F  在 n  方向的投影,即 rotF  在 z 轴方向的分量乘以(-1))  = 1 2 3 S x dS  +  = 4 2 2 2 3 x y x dxdy   = 2 0 3 2 2 0 d 3r cos dr    2 12 4 1 4 3 4 4 =    =