Determine(a)a rotacional e(b)a divergente do campo vetorial.
\mathbf{F}(x, y, z)=e^{x z} \mathbf{i}-2 e^{y z} \mathbf{j}+3 x e^{y} \mathbf{k}

Passo 1

O enunciado pede o rotacional e a divergente do campo
\mathbf{F}(x,y,z)=e^{x z}\,\mathbf{i}-2e^{y z}\,\mathbf{j}+3x e^{y}\,\mathbf{k}.
Estamos em cálculo vetorial: definimos as componentes
\displaystyle P(x,y,z)=e^{xz},\;Q(x,y,z)=-2e^{yz},\;R(x,y,z)=3x e^y
e usamos as fórmulas
\nabla\times\mathbf{F} =\bigl(\partial_yR-\partial_zQ\bigr)\,\mathbf{i} +\bigl(\partial_zP-\partial_xR\bigr)\,\mathbf{j} +\bigl(\partial_xQ-\partial_yP\bigr)\,\mathbf{k},
\nabla\cdot\mathbf{F} =\partial_xP+\partial_yQ+\partial_zR.

Passo 2

Calculemos \partial_yR e \partial_zQ:
\partial_yR=\partial_y\bigl(3x e^y\bigr)=3x e^y, \qquad \partial_zQ=\partial_z\bigl(-2e^{yz}\bigr)=-2\,y\,e^{yz}.
Logo a componente \mathbf{i} do rotacional é
\partial_yR-\partial_zQ =3x e^y-(-2y e^{yz}) =3x e^y+2y e^{yz}.

Passo 3

Agora \partial_zP e \partial_xR:
\partial_zP=\partial_z\bigl(e^{xz}\bigr)=x e^{xz}, \qquad \partial_xR=\partial_x\bigl(3x e^y\bigr)=3 e^y.
Portanto a componente \mathbf{j} é
\partial_zP-\partial_xR =x e^{xz}-3 e^y.

Passo 4

Por fim \partial_xQ e \partial_yP:
\partial_xQ=\partial_x\bigl(-2e^{yz}\bigr)=0, \qquad \partial_yP=\partial_y\bigl(e^{xz}\bigr)=0,
logo a componente \mathbf{k} vale
\partial_xQ-\partial_yP=0-0=0.
Assim concluímos
\nabla\times\mathbf{F} =(3x e^y+2y e^{yz})\,\mathbf{i} +(x e^{xz}-3 e^y)\,\mathbf{j} +0\,\mathbf{k}.

Passo 5

Para a divergente, calculamos
\partial_xP=\partial_x(e^{xz})=z e^{xz}, \quad \partial_yQ=\partial_y(-2e^{yz})=-2\,z\,e^{yz}, \quad \partial_zR=\partial_z(3x e^y)=0
e somamos:
\nabla\cdot\mathbf{F} =z e^{xz}-2z e^{yz}+0 =z\bigl(e^{xz}-2e^{yz}\bigr).

Resposta

(a) \displaystyle\nabla\times\mathbf{F} =(3x e^y+2y e^{yz})\,\mathbf{i}+(x e^{xz}-3 e^y)\,\mathbf{j}.
(b) \displaystyle\nabla\cdot\mathbf{F} =z\bigl(e^{xz}-2e^{yz}\bigr).

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