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(√x^2+ y^2)，求二重积分。

164 设f(x,y)有二阶连续偏导数, (x,y)=f((e)^xy,(x)^2+(y)^2), 且 f(x,y)=1-x-y+-|||-(sqrt ({(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(

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

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

设f(x,y)= ^4+{y)^2},(x)^2+(y)^2neq 0 0,(x)^2+(y)^2=0 .处是否连续？设,试问在点处是否连续？

11.设 =f((x)^2+(y)^2) ,其中f具有二阶导数,求 dfrac ({a)^2z}(d{x)^2} -dfrac ({partial )^2z}(

设:(x)^2+(y)^2leqslant 1,则二重积分:(x)^2+(y)^2leqslant 1设,则二重积分

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（）。

设 (x,y)=dfrac ({x)^2+(y)^2}({e)^xy+xysqrt ({x)^2+(y)^2}} ,则 (f)_(x)(1,0)= __ _.

讨论下列函数的连续性:-|||-^2+(y)^2neq 0,-|||-f(x,y)= ) (y)^2ln ((x)^2+(y)^2) 0,=0;