例1 计算 (int )_(1)sqrt ({x)^2+(y)^2}ds, 其中L为圆周 ^2+(y)^2=4x.

[题目]设L为球面 ^2+(y)^2+(z)^2=(a)^22 被平面 x+y+z=0 所-|||-截的圆周,则 (int )_(1)((x)^2+(y)^2)

求指导本题解题过程，谢谢您！2.曲线积分 =(int )_(L)^1((x)^2+(y)^2)ds (其中L是圆周: ^2+(y)^2=9 的值为 __ _。求

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}

例3.2 计算曲面积分 (int )_(dfrac {1)(2)}^x(y)_(x)zdxdy 其中∑是球面 ^2+(y)^2+(z)^2=1(xgeqslan

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

计算(int )_(0)^1dx(int )_(1-x)^sqrt (1-{x^2)}dfrac (x+y)({x)^2+(y)^2}dy=-|||-dv=__

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

例1 计算 iint xydxdy, 其中D为由下列双纽线所围成:(1) (({x)^2+(y)^2)}^2=-|||-((x)^2-(y)^2); (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

4.计算下列曲面积分:-|||-(3) int [ ((x+y))^2+(z)^2+2yz] dS, 其中∑是球面 ^2+(y)^2+(z)^2=2x+2z;