Signals not enabled for storage cannot be part of the Storage Qualifier condition
(n|N) die "Not saved, result left in file new" ;; esac done :<<'auxfiles' begin-base64 644 - UEsDBBQACAAIAMOCfFQAAAAAAAAAAAECAAAJABwAUkVBRE1FLm1kVVQJAAOO xEFihsVBYnV4CwABBOgDAAAE6AMAAEWRwWrDMAyG734KwS4tNFnHbj1t0JJL GTu0bDB2cBylMXXsYMvzAn34yUm7BhPwL+nTL/kBTDyJC+TvjCNcoMGgvB5I...
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设du=(2x+y)dx+(x+2y)dy,求u(x,y)解:本题的du较简单,不需要利用第二类曲线积分求。将du恒等变形,得du=2xdx+2ydy+(xdy+ydx),于是由微分的基本公式,可得u(x,y)=,C为任意常数。这里 M=___,m=___,N=___,n=___,P=___,p=___,q=___ 参考答案: ...
26、Vxdy=4ydx+(4x+1)dy= d=dx+dy=-Vydx+Vxdy= 4ydx+(4x+1)dy=4xy+y6-2 平面不可压缩流体速度分布:Vx=x2-y2+x; Vy=-(2xy+y).(1) 流动满足连续性方程否? (2) 势函数、流函数存在否? (3)求、 .解:(1)由于+=2x1(2x1)0,故该流动满足连续性方程,流动存在.(2)由z=()=0, 故流...
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t0 (10) 2 x cos xdy = 0 . (11)已知曲线 L 的方程为 y 1 x {x[1,1]}, 起点是 (1,0), 终点是 (1, 0), 则曲线积分 L xydx x2dy = . (12) 设 {(x, y, z) | x2 y2 z 1}, 则的形心的竖坐标 z= . (13)设α1 (1, 2, 1, 0)T , α2 (1,1, 0, 2)T , α...
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