18设力F的模 |F|=dfrac (1)(sqrt {{x)^2+(y)^2}}, F的方向与 =-yi+x(p)^2 相同,则在力F的作用下,质点沿曲线 :dfrac ({x)^2}({a)^2}+dfrac ({y)^2}({b)^2}=1-|||-正向绕行一周,力F所做的功可用曲线积分标识为().-|||-○

(9)设 f(x,y)= dfrac (1)({({x)^2+(y)^2)}^2},1leqslant xleqslant 3, dfrac (sqrt {3)

11.设 (x)=dfrac (x)(sqrt {1+{x)^2}} 求 [ f(x)] , f[ f(x)] .

设=dfrac (y)(f({x)^2-(y)^2)},其中f为可导函数.验证=dfrac (y)(f({x)^2-(y)^2)}.设,其中f为可导函数.验证.

设(x)+(x)^2f(dfrac (1)(x))=dfrac ({x)^2+2x}(x+1) 求f(x)设

设 =f(xy,(x)^2+(y)^2), 其中 f 可微,则 dfrac (partial z)(partial x)= __

数值求积公式(int )_(-1)^1f(x)dxapprox dfrac (2)(3)[ f(-dfrac (1)(sqrt {2)})+f(0)+f(dfr

1.设 (x)=dfrac (1)(1-{x)^2}, 求 (-x),f[ f(x)] ,f[ dfrac (1)(f(x))]

已知函数 z=f(x,y) 连续且满足 lim _(xarrow 1)dfrac (f(x,y)-x+2y+2)(sqrt {{(x-1))^2+(y)^2}}

证明:函数-|||-f(x,y)= ((x)^2+(y)^2)sin dfrac (1)(sqrt {{x)^2+(y)^2}}, ^2+(y)^2neq 0,

九、设f(x,y)在 ^2+(y)^2leqslant 1 上二次连续可微,且满足 dfrac ({partial )^2f}(partial {x)^2}+d