We are given ϕ(x,y,z,)=2x2y−xzn3\phi(x,y,z,) = 2 x^2 y - x z n^3ϕ(x,y,z,)=2x2y−xzn3.
The Laplacian of ϕ(x,y,z)\phi(x,y,z)ϕ(x,y,z) in Euclidean coordinates is △ϕ(x,y,z)=∂2ϕ∂x2+∂2ϕ∂y2+∂2ϕ∂z2\triangle \phi(x,y,z) = \frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2} + \frac{\partial^2 \phi}{\partial z^2}△ϕ(x,y,z)=∂x2∂2ϕ+∂y2∂2ϕ+∂z2∂2ϕ. Let us calculate the derivatives:
∂ϕ∂x=4xy−zn3\frac{\partial \phi}{\partial x} = 4 x y - z n^3∂x∂ϕ=4xy−zn3, ∂2ϕ∂x2=4y\frac{\partial^2 \phi}{\partial x^2} = 4 y∂x2∂2ϕ=4y,
∂ϕ∂y=2x2\frac{\partial \phi}{\partial y} = 2 x^2∂y∂ϕ=2x2, ∂2ϕ∂y2=0\frac{\partial^2 \phi}{\partial y^2} = 0∂y2∂2ϕ=0,
∂ϕ∂z=−xn3\frac{\partial \phi}{\partial z} = - x n^3∂z∂ϕ=−xn3, ∂2ϕ∂z2=0\frac{\partial^2 \phi}{\partial z^2} = 0∂z2∂2ϕ=0.
Hence, △ϕ(x,y,z)=4y\triangle \phi(x,y,z) = 4 y△ϕ(x,y,z)=4y.
