已知圆x^2+(y+2)^2=r^2(r>0)上到直线y=sqrt(3)x+2的距离为1的点有且仅有2个，则r的取值范围是( )

4.若圆x^2+(y+2)^2=r^2(r>0)上到直线y=sqrt(3)x+2的距离为1的点有且仅有2个，则r的取值范围是A. (0,1)B. (1,3)C.

7.设 (x,y)=dfrac (1)(xy),r=sqrt ({x)^2+(y)^2} _(1)= (x,y)|(x,y)in {R)^2 dfrac (1)

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

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}

, ^2+(y)^2leqslant (R)^2,-|||-其他,-|||-求:(1)常数c; (2) {X)^2+(Y)^2leqslant (r)^2}

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

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

已知函数 z=f(x,y) 连续且满足 lim _(xarrow 1)dfrac (f(x,y)-x+2y+2)(sqrt {{(x-1))^2+(y)^2}}

4.设∑表示圆柱面 ^2+(y)^2=(R)^2 介于 z=0 和 z=2 之间的部分,则曲面积分-|||-int (dfrac (1)(sqrt {{x)^2