设数量场 =ln sqrt ({x)^2+(y)^2+(z)^2} 则rot(gradu) (1,0,1)1 ,= () .-|||-(A) dfrac (1)(4) (B)0 (C)(0,0,0) (D)(1,0,1)

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

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

1.设 ^2+(y)^2+(z)^2-z=0, 求 dfrac ({a)^2z}(a{y)^2}

设 f(x,y)= sin dfrac (1)({x)^2+(y)^2} , ^2+(y)^2neq 0,-|||-0, ^2+(y)^2=0,-|||-考察函

设函数y=f(x)由方程(y)^2+(y)^2ln x-4=0所确定,则(y)^2+(y)^2ln x-4=0= ( )(y)^2+(y)^2ln x-4=0(

12.设方程 +sqrt ({x)^2+(y)^2+(z)^2}=sqrt (2) 确定了函数 =z(x,y), 则z(x,y)-|||-在点 (1,0,-1)

(5)设 (x,y)=ln (x+dfrac (y)(2x)) ,则 _(y)(1,0)= () .-|||-(A)1 (B) dfrac (1)(2) (C)

曲线 ) (x)^2-(y)^2+(z)^2=0 z=1 .曲线在坐标面上的投影曲线方程为______A.B.C.D.

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