设r=√(x-x ) (}^2+{(y-y))^2+((z-z))^2为源点x'到场点X的距离，r的方向规定为从源点指向场点。r=√(x-x ) (}^2+{(y-y))^2+((z-z))^2

从源点(即电荷电流分布点)x到场点x的距离r，以及矢径r分别为=sqrt ({(x-x))^2+((y-y))^2+((z-z))^2}-|||-=(x-x)(

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

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I为空间曲线 ) (x)^2+(y)^2+(z)^2=(R)^2 x+y+z=0-|||-B πR 2-|||-C 2πR 3-|||-D 2πR 2

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)

函数(z)=(x)^2+(y)^2i ( ).A.仅在(z)=(x)^2+(y)^2i上解析;B.在除(z)=(x)^2+(y)^2i之外的复平面上

16.设函数 z=z(x,y) 由方程 ^2+(y)^2+(z)^2-6z=0 确定,求 dfrac ({sigma )^2z}(sigma x{U)_(y)}

[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)

设 ^2+(y)^2+(z)^2-4z=0, 求 dfrac ({a)^2z}(q{x)^2}

1.设 (x,y,z)=dfrac (z)({x)^2+(y)^2}, 则 df(1,2,1)= __