点(1,-1,1)在下面的某个曲面上,该曲面是( )。
A: x^2 +y^2=z
B: x^2 +y^2 — 2z =0
C: z=ln(x^2+y^2)
D: x^2 +y^2+2z =0
A: x^2 +y^2=z
B: x^2 +y^2 — 2z =0
C: z=ln(x^2+y^2)
D: x^2 +y^2+2z =0
举一反三
- 4.已知二元函数$z(x,y)$满足方程$\frac{{{\partial }^{2}}z}{\partial x\partial y}=x+y$,并且$z(x,0)=x,z(0,y)={{y}^{2}}$,则$z(x,y)=$( ) A: $\frac{1}{2}({{x}^{2}}y-x{{y}^{2}})+{{y}^{2}}+x$ B: $\frac{1}{2}({{x}^{2}}{{y}^{2}}+xy)+{{y}^{2}}+x$ C: ${{x}^{2}}{{y}^{2}}+{{y}^{2}}+x$ D: $\frac{1}{2}({{x}^{2}}y+x{{y}^{2}})+{{y}^{2}}+x$
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