设=sqrt ({x)^2+(y)^2+(z)^2} 则|div(grad)|(1,0,1)= () .-|||-(A) -sqrt (2) (B) sqrt (2) (C)0

设数量场 =ln sqrt ({x)^2+(y)^2+(z)^2} 则rot(gradu) (1,0,1)1 ,= () .-|||-(A) dfrac (1)

12.设方程 +sqrt ({x)^2+(y)^2+(z)^2}=sqrt (2) 确定了函数 =z(x,y), 则z(x,y)-|||-在点 (1,0,-1)

已知Σ为锥面=sqrt ({x)^2+(y)^2}在柱体=sqrt ({x)^2+(y)^2}内的部分,则曲面积分=sqrt ({x)^2+(y)^2}

设函数 (x,y)=1-dfrac (cos sqrt {{x)^2+(y)^2}}(tan ({x)^2+(y)^2)} ,则当定设函数 (x,y)=1-df

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

10、曲面 =sqrt ({x)^2+(y)^2} 被 ^2+(y)^2=1 所截部分的-|||-面积为 () .-|||-(A)π;(B) √2π; (C)2

12.试用向量证明不等式:-|||-sqrt ({{a)_(1)}^2+({a)_(2)}^2+({a)_(3)}^2}sqrt ({{b)_(1)}^2+({

[例5] 积分 (int )_(0)^2dx(int )_(0)^sqrt (2x-{x^2)}sqrt ({x)^2+(y)^2}dy= __

4.设∑表示圆柱面 ^2+(y)^2=(R)^2 介于 z=0 和 z=2 之间的部分,则曲面积分-|||-int (dfrac (1)(sqrt {{x)^2

10.设总体 sim N(0,1), X1,X2,X3是来自X的样本,则 =dfrac (sqrt {3)X}(sqrt {{{X)_(1)}^2+({X)_(