Answer to Question #96504 in Vector Calculus for Olajide Olaitan

Question #96504
Find the laplacian of the function (phi(x,y,z)=2x^2y-xzn^3
1
Expert's answer
2019-10-16T13:21:57-0400

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".


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