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} (A)=sqrt ({x)^2+(y)^2} (B)=sqrt ({x)^2+(y)^2} (C)2 (D)0

已知(x,y)=(e)^x+(y-2)arcsin sqrt ({x)^2+(y)^2}，求(x,y)=(e)^x+(y-2)arcsin sqrt ({x)^

[题目]-|||-求锥面 =sqrt ({x)^2+(y)^2} 被柱面-|||-^2+(y)^2=x 所割下部分的曲面面积.

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

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

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

(6) arctan dfrac (y)(x)=ln sqrt ({x)^2+(y)^2}

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

[题目]已知函数 =ln (x+sqrt ({a)^2+(x)^2}) 求y`.

设D: leqslant (x)^2+(y)^2, =dfrac (dxdy)(sqrt {{x)^2+(y)^2}}=()leqslant yleqslant

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