Determine o rotacional e o divergente do campo indicado:
\mathrm{F}(\mathrm{x}, \mathrm{y}, \mathrm{z})=\mathrm{x}^{2} \operatorname{sen}(\mathrm{yz}) \mathbf{i}+\mathrm{z} \cos \left(\mathrm{xz}^{3}\right) \mathbf{j}+\mathrm{ye}^{5 \mathrm{xy}} \mathbf{k}

Passo 1

O enunciado pede para calcular a divergente e o rotacional do campo de vetor
F(x,y,z)=x^{2}\sin(yz)\,\mathbf{i}+z\cos\bigl(xz^{3}\bigr)\,\mathbf{j}+y\,e^{5xy}\,\mathbf{k}.
O conteúdo usado é o operador nabla em coordenadas cartesianas, com
\div F = \frac{\partial F_x}{\partial x}+\frac{\partial F_y}{\partial y}+\frac{\partial F_z}{\partial z},
\nabla\times F = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\\ \partial/\partial x & \partial/\partial y & \partial/\partial z\\\ F_x & F_y & F_z \end{vmatrix}.

Passo 2

Calcular a divergente de F:
\div F =\frac{\partial}{\partial x}\!\bigl(x^{2}\sin(yz)\bigr) +\frac{\partial}{\partial y}\!\bigl(z\cos(xz^{3})\bigr) +\frac{\partial}{\partial z}\!\bigl(y\,e^{5xy}\bigr).
  1. \displaystyle \frac{\partial}{\partial x}\bigl(x^{2}\sin(yz)\bigr) =2x\sin(yz).
  2. \displaystyle \frac{\partial}{\partial y}\bigl(z\cos(xz^{3})\bigr)=0 (não depende de y).
  3. \displaystyle \frac{\partial}{\partial z}\bigl(y\,e^{5xy}\bigr)=0 (não depende de z).
Logo
\div F = 2x\sin(yz).

Passo 3

Calcular o componente x do rotacional:
(\nabla\times F)_x =\frac{\partial F_z}{\partial y}-\frac{\partial F_y}{\partial z}.
  1. \displaystyle \frac{\partial}{\partial y}\bigl(y\,e^{5xy}\bigr) =e^{5xy}+y\cdot(5x)e^{5xy} =e^{5xy}(1+5xy).
  2. \displaystyle \frac{\partial}{\partial z}\bigl(z\cos(xz^{3})\bigr) =\cos(xz^{3})+z\bigl(-\sin(xz^{3})\cdot 3xz^{2}\bigr) =\cos(xz^{3})-3xz^{3}\sin(xz^{3}).
Portanto
(\nabla\times F)_x =e^{5xy}(1+5xy)\;-\;\bigl[\cos(xz^{3})-3xz^{3}\sin(xz^{3})\bigr].

Passo 4

Calcular o componente y do rotacional:
(\nabla\times F)_y =\frac{\partial F_x}{\partial z}-\frac{\partial F_z}{\partial x}.
  1. \displaystyle \frac{\partial}{\partial z}\bigl(x^{2}\sin(yz)\bigr) =x^{2}\cos(yz)\,\cdot y =x^{2}y\cos(yz).
  2. \displaystyle \frac{\partial}{\partial x}\bigl(y\,e^{5xy}\bigr) =y\cdot(5y)e^{5xy} =5y^{2}e^{5xy}.
Logo
(\nabla\times F)_y =x^{2}y\cos(yz)\;-\;5y^{2}e^{5xy}.

Passo 5

Calcular o componente z do rotacional:
(\nabla\times F)_z =\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}.
  1. \displaystyle \frac{\partial}{\partial x}\bigl(z\cos(xz^{3})\bigr) =z\bigl(-\sin(xz^{3})\cdot z^{3}\bigr) =-z^{4}\sin(xz^{3}).
  2. \displaystyle \frac{\partial}{\partial y}\bigl(x^{2}\sin(yz)\bigr) =x^{2}\cos(yz)\,\cdot z =x^{2}z\cos(yz).
Portanto
(\nabla\times F)_z =-z^{4}\sin(xz^{3})\;-\;x^{2}z\cos(yz).

Passo 6

Juntando todos os componentes, obtemos
\nabla\times F =\bigl[e^{5xy}(1+5xy)-\cos(xz^{3})+3xz^{3}\sin(xz^{3})\bigr]\,\mathbf{i}
\quad+\bigl[x^{2}y\cos(yz)-5y^{2}e^{5xy}\bigr]\,\mathbf{j} +\bigl[-\,z^{4}\sin(xz^{3})-x^{2}z\cos(yz)\bigr]\,\mathbf{k}.

Resposta

\displaystyle \div F = 2x\sin(yz) \displaystyle \nabla\times F =\bigl[e^{5xy}(1+5xy)-\cos(xz^{3})+3xz^{3}\sin(xz^{3})\bigr]\,\mathbf{i} +\bigl[x^{2}y\cos(yz)-5y^{2}e^{5xy}\bigr]\,\mathbf{j} +\bigl[-z^{4}\sin(xz^{3})-x^{2}z\cos(yz)\bigr]\,\mathbf{k}.

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