Determine (a) o rotacional e (b) o divergente do campo vetorial.
\mathbf{F}(x, y, z)=e^{x} \operatorname{sen} y \mathbf{i}+e^{x} \cos y \mathbf{j}+z \mathbf{k}

Passo 1

O enunciado pede o rotacional e o divergente do campo vetorial
F(x,y,z)=e^xsin y,i+e^xcos y,j+z,k.
O conteúdo é cálculo vetorial: utilizaremos as definições
nablatimesF (rotacional) e nablacdotF (divergente) e calcularemos as derivadas parciais necessárias.

Passo 2

Lembrando a fórmula do rotacional em coordenadas cartesianas, podemos usar o determinante
nabla times F = i & j & k partialpartial x & partialpartial y & partialpartial z e^xsin y & e^xcos y & z
Isso gera, componente a componente:
(nabla times F)x =partial Fzpartial y-partial Fypartial z, (nabla times F)y =partial Fxpartial z-partial Fzpartial x, (nabla times F)z =partial Fypartial x-partial Fxpartial y

Passo 3

Calculemos cada componente:
(nabla times F)x =partial zpartial y-partial (e^xcos y)partial z =0-0=0,
(nabla times F)y =partial (e^xsin y)partial z-partial zpartial x =0-0=0,
(nabla times F)z =partial (e^xcos y)partial x-partial (e^xsin y)partial y =e^xcos y - e^xcos y=0
Logo nablatimesF=0

Passo 4

Para o divergente, usamos a definição
nablacdotF =partial Fxpartial x +partial Fypartial y +partial Fzpartial z
As derivadas parciais são
partial Fxpartial x =partial (e^xsin y)partial x =e^xsin y, partial Fypartial y =partial (e^xcos y)partial y =-,e^xsin y, partial Fzpartial z =partial zpartial z =1

Passo 5

Somando as três parcelas obtemos
nablacdotF =e^xsin y - e^xsin y + 1 = 1

Resposta

(a) nablatimesF = 0
(b) nablacdotF = 1

Exercícios relacionados