求旋转曲面=(x)^2+(y)^2在点=(x)^2+(y)^2处的法线方程A.=(x)^2+(y)^2B.=(x)^2+(y)^2C.=(x)^2+(y)^2D.=(x)^2+(y)^2

( A ) = (x,y,z)|{x)^2+(y)^2+(z)^2=(a)^2,zgeqslant 0} ( B ) = (x,y,z)|{x)^2+(y)^

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}

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

dfrac (dy)(dx)=(x)^2+(y)^2 B . dfrac (dy)(dx)=(x)^2+(y)^2 C .dfrac (dy)(dx)=(x)^

设=u(√x^2+ y^2)有二阶连续偏导数，且满足=u(√x^2+ y^2)，=u(√x^2+ y^2)。（I）求=u(√x^2+ y^2)；（II）求=u(

函数f(x,y)= dfrac (xy)({x)^2+(y)^2},(x)^2+(y)^2neq 0-|||-0, ^2+(y)^2=0在点（0，0）处（）。

球面^2+(y)^2+(z)^2+4x+6y+2z+10=0的球心坐标为A.^2+(y)^2+(z)^2+4x+6y+2z+10=0B.^2+(y)^2+(z)

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

设(x,y,z)=(x)^2+(y)^3+z,求(x,y,z)=(x)^2+(y)^3+z,在点(x,y,z)=(x)^2+(y)^3+z,处沿方向(x,y,z

设函数=f(√x^2+ y^2)，其中=f(√x^2+ y^2)当=f(√x^2+ y^2)时具有二阶连续导数，并且满足=f(√x^2+ y^2)，平面区域=f