(3)曲线 ) 与圆柱面 ^2+{y)^2=4 的交点为 () .-|||-(A) (sqrt (2),sqrt (2),pm 2) (B) (sqrt (2),sqrt (2),2)-|||-(C) (sqrt (2),sqrt (2),-2) (D) (sqrt (2),sqrt (2),pm 1)

2.根式计算.-|||-(1) sqrt (dfrac {4)(25)} ;-|||-(2) sqrt ({11)^2+((4sqrt {3))}^2}-|||

已知x=(sqrt(3)-sqrt(2))/(sqrt(3)+sqrt(2))，y=(sqrt(3)+sqrt(2))/(sqrt(3)-sqrt(2))，求3

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}

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

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

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

(int )_(-sqrt {2)}^sqrt (2)sqrt (8-2{y)^2}dy ;

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

计算：（1）（sqrt(5)）2；（2）（-sqrt(0.2)）2；（3）（sqrt((2)/(7))）2；（4）（5sqrt(5)）2；（5）sqrt(（-1

2.求曲线 ^dfrac (2{3)}+(y)^dfrac (2{3)}=(a)^dfrac (2{3)} 在点 (dfrac (sqrt {2)}(4)a,d