(2)数列 √2, sqrt (2+sqrt {2)} , √(2+√2+√2),··· 的极限存在

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}

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

根据数列极限的定义证明:-|||-lim dfrac (sqrt {{n)^2+(a)^2}}(n)=1

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

sqrt ({11)^2+((4sqrt {3))}^2}

(7) sqrt ({11)^2+((4sqrt {3))}^2}

(3)曲线 ) 与圆柱面 ^2+{y)^2=4 的交点为 () .-|||-(A) (sqrt (2),sqrt (2),pm 2) (B) (sqrt (

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

+dfrac (1)(sqrt {{n)^2+(n)^2}})= )=______.______.

=ln (x+sqrt ({a)^2+(x)^2});