Determine o rotacional e o divergente do campo vetorial indicado.
\boldsymbol{F}(x, y, z)=x^{2} \operatorname{sen}(y z) \boldsymbol{i}+z \cos \left(x z^{3}\right) \boldsymbol{j}+y e^{5 x y} \boldsymbol{k}

Passo 1

O enunciado dá o campo vetorial
\boldsymbol{F}(x,y,z)=x^{2}\sin(yz)\,\boldsymbol{i}+z\cos\bigl(xz^{3}\bigr)\,\boldsymbol{j}+y\,\mathrm{e}^{5xy}\,\boldsymbol{k}
e pede o divergente \nabla\cdot\boldsymbol{F} e o rotacional \nabla\times\boldsymbol{F}. Vamos usar as definições de cálculo vetorial.

Passo 2

A definição de divergente de \boldsymbol{F}=(F_{x},F_{y},F_{z}) é
\nabla\cdot\boldsymbol{F} =\frac{\partial F_{x}}{\partial x} +\frac{\partial F_{y}}{\partial y} +\frac{\partial F_{z}}{\partial z}.
Calculamos cada derivada parcial:

Passo 3

\frac{\partial F_{x}}{\partial x} =\frac{\partial}{\partial x}\bigl(x^{2}\sin(yz)\bigr) =2x\sin(yz),
\frac{\partial F_{y}}{\partial y} =\frac{\partial}{\partial y}\bigl(z\cos(xz^{3})\bigr) =0,
\frac{\partial F_{z}}{\partial z} =\frac{\partial}{\partial z}\bigl(y\,\mathrm{e}^{5xy}\bigr) =0.

Passo 4

Somando, obtemos o divergente
\nabla\cdot\boldsymbol{F} =2x\sin(yz).

Passo 5

A definição de rotacional de \boldsymbol{F}=(F_{x},F_{y},F_{z}) é
\nabla\times\boldsymbol{F} =\Bigl(\frac{\partial F_{z}}{\partial y}-\frac{\partial F_{y}}{\partial z}\Bigr)\boldsymbol{i} +\Bigl(\frac{\partial F_{x}}{\partial z}-\frac{\partial F_{z}}{\partial x}\Bigr)\boldsymbol{j} +\Bigl(\frac{\partial F_{y}}{\partial x}-\frac{\partial F_{x}}{\partial y}\Bigr)\boldsymbol{k}.
Vamos calcular cada termo.

Passo 6

Para o componente \boldsymbol{i}:
\frac{\partial F_{z}}{\partial y} =\frac{\partial}{\partial y}\bigl(y\,\mathrm{e}^{5xy}\bigr) =\mathrm{e}^{5xy}+5xy\,\mathrm{e}^{5xy} =\mathrm{e}^{5xy}(1+5xy),
\frac{\partial F_{y}}{\partial z} =\frac{\partial}{\partial z}\bigl(z\cos(xz^{3})\bigr) =\cos(xz^{3})-3xz^{3}\sin(xz^{3}).
Logo,
(\nabla\times\boldsymbol{F})_{i} =\mathrm{e}^{5xy}(1+5xy) -\bigl[\cos(xz^{3})-3xz^{3}\sin(xz^{3})\bigr].

Passo 7

Para o componente \boldsymbol{j}:
\frac{\partial F_{x}}{\partial z} =\frac{\partial}{\partial z}\bigl(x^{2}\sin(yz)\bigr) =x^{2}y\cos(yz),
\frac{\partial F_{z}}{\partial x} =\frac{\partial}{\partial x}\bigl(y\,\mathrm{e}^{5xy}\bigr) =5y^{2}\,\mathrm{e}^{5xy}.
Logo,
(\nabla\times\boldsymbol{F})_{j} =x^{2}y\cos(yz)-5y^{2}\,\mathrm{e}^{5xy}.

Passo 8

Para o componente \boldsymbol{k}:
\frac{\partial F_{y}}{\partial x} =\frac{\partial}{\partial x}\bigl(z\cos(xz^{3})\bigr) =-z^{4}\sin(xz^{3}),
\frac{\partial F_{x}}{\partial y} =\frac{\partial}{\partial y}\bigl(x^{2}\sin(yz)\bigr) =x^{2}z\cos(yz).
Logo,
(\nabla\times\boldsymbol{F})_{k} =-z^{4}\sin(xz^{3})-x^{2}z\cos(yz).

Passo 9

Reunindo, o rotacional fica
\nabla\times\boldsymbol{F} =\bigl[\mathrm{e}^{5xy}(1+5xy)-\cos(xz^{3})+3xz^{3}\sin(xz^{3})\bigr]\boldsymbol{i}
\qquad\quad +\bigl[x^{2}y\cos(yz)-5y^{2}\,\mathrm{e}^{5xy}\bigr]\boldsymbol{j} +\bigl[-z^{4}\sin(xz^{3})-x^{2}z\cos(yz)\bigr]\boldsymbol{k}.

Resposta

Divergente:
\nabla\cdot\boldsymbol{F}=2x\sin(yz).
Rotacional:
\nabla\times\boldsymbol{F} =\bigl[\mathrm{e}^{5xy}(1+5xy)-\cos(xz^{3})+3xz^{3}\sin(xz^{3})\bigr]\boldsymbol{i}
\quad +\bigl[x^{2}y\cos(yz)-5y^{2}\,\mathrm{e}^{5xy}\bigr]\boldsymbol{j} +\bigl[-z^{4}\sin(xz^{3})-x^{2}z\cos(yz)\bigr]\boldsymbol{k}.

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