Determine Laplaciano dos campos escalares abaixo:
w=10 r sen^2(theta) cos (phi).

Passo 1

O enunciado pede o Laplaciano do escalar
w(r,theta,phi) = 10,r,sin^2(theta),cos(phi)
em coordenadas esféricas. Vamos usar a expressão do Laplaciano em esféricas e calcular cada termo.

Passo 2

Em coordenadas esféricas, para uma função f(r,theta,phi) vale
nabla^2 f = 1r^2partialpartial r!l(r^2partial fpartial rr) + 1r^2sinthetapartialpartial theta!l(sinthetapartial fpartial thetar) + 1r^2sin^2thetapartial^2 fpartial phi^2

Passo 3

Cálculo da parte radial:
partial wpartial r =10,sin^2theta,cosphi,
1r^2partialpartial r!l(r^2partial wpartial rr) =1r^2partialpartial r!l(10,r^2sin^2theta,cosphir) =1r^2l(20,r,sin^2theta,cosphir) =20,sin^2theta,cosphir

Passo 4

Cálculo da parte polar ( theta):
partial wpartial theta =10,rcdot2sinthetacostheta;cosphi =20,r,sinthetacosthetacosphi,
sinthetapartial wpartial theta =20,r,sin^2thetacosthetacosphi,
partialpartial theta!l(sinthetapartial wpartial thetar) =20,r,cosphi,ddthetal(sin^2thetacosthetar) =20,r,cosphi,sintheta,(3cos^2theta-1),
1r^2sinthetapartialpartial theta!l(sinthetapartial wpartial thetar) =20,r,cosphi,sintheta,(3cos^2theta-1)r^2sintheta =20,(3cos^2theta-1)cosphir

Passo 5

Cálculo da parte azimutal ( phi):
partial wpartial phi =10,r,sin^2theta,(-sinphi) =-10,r,sin^2theta,sinphi,
partial^2 wpartial phi^2 =partialpartial phil(-10,r,sin^2theta,sinphir) =-10,r,sin^2theta,cosphi,
1r^2sin^2thetapartial^2 wpartial phi^2 =-10,r,sin^2theta,cosphir^2sin^2theta =-10,cosphir

Passo 6

Somando os três termos:
nabla^2 w =20,sin^2theta,cosphir +20,(3cos^2theta-1)cosphir -10,cosphir
Como sin^2theta=1-cos^2theta, a soma das constantes dá
20(1-cos^2theta)-20-10 = -10+40cos^2theta,
logo
nabla^2 w =(-10+40cos^2theta)cosphir =10,(4cos^2theta-1),cosphir

Resposta

nabla^2 w =10,(4cos^2theta-1),cosphir

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