Question

Find the laplacian of the function (phi(x,y,z)=2x^2y-xzn^3

Answer 1

We are given $\phi(x,y,z,) = 2 x^2 y - x z n^3$.

The Laplacian of $\phi(x,y,z)$ in Euclidean coordinates is $\triangle \phi(x,y,z) = \frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2} + \frac{\partial^2 \phi}{\partial z^2}$. Let us calculate the derivatives:

$\frac{\partial \phi}{\partial x} = 4 x y - z n^3$, $\frac{\partial^2 \phi}{\partial x^2} = 4 y$,

$\frac{\partial \phi}{\partial y} = 2 x^2$, $\frac{\partial^2 \phi}{\partial y^2} = 0$,

$\frac{\partial \phi}{\partial z} = - x n^3$, $\frac{\partial^2 \phi}{\partial z^2} = 0$.

Hence, $\triangle \phi(x,y,z) = 4 y$.