Determine Laplaciano dos campos escalares abaixo:
$u=\rho^{2} z \cos (2 \varphi);

Passo 1

O enunciado pede o laplaciano do campo escalar em coordenadas cilíndricas
u(rho,phi,z)=rho^2,z,cos(2phi)
O laplaciano em coordenadas cilíndricas é
nabla^2u =1rhopartialpartialrho!l(rho,partial upartialrhor) +1rho^2partial^2upartialphi^2 +partial^2upartial z^2
Vamos calcular termo a termo.

Passo 2

Cálculo do termo radial
partial upartialrho =partialpartialrhol(rho^2zcos(2phi)r) =2rho,zcos(2phi),
rho,partial upartialrho =2rho^2zcos(2phi),
partialpartialrho!l(rho,partial upartialrhor) =partialpartialrhol(2rho^2zcos(2phi)r) =4rho,zcos(2phi),
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1rhopartialpartialrho!l(rho,partial upartialrhor) =4,zcos(2phi)

Passo 3

Cálculo do termo angular
partial upartialphi =rho^2z,ddphil(cos(2phi)r) =-2rho^2z,sin(2phi),
partial^2upartialphi^2 =partialpartialphil(-2rho^2zsin(2phi)r) =-4rho^2zcos(2phi),
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1rho^2partial^2upartialphi^2 =-4,zcos(2phi)

Passo 4

Cálculo do termo axial
partial upartial z =rho^2cos(2phi), partial^2upartial z^2 =0

Passo 5

Somando todos os termos do laplaciano
nabla^2u =4,zcos(2phi);+;l(-4,zcos(2phi)r);+;0 =0

Resposta

nabla^2u = 0

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