\begin{tabular}{l} Dado el campo vectorial $$ \mathbf{F}(x, y, z)=\left(y^{2}+3 z ight) \mathbf{i}+(2 x+z) \mathbf{j}+\left(x^{2}-y ight) \mathbf{k} $$ \ ¿cuál es el rotacional $$
abla imes F $$ ? \ \begin{tabular}{l}$$ 0-1 \mathbf{i}-(2 x-3) \mathbf{j}+(2-2 y) \mathbf{k} $$ \ $$ \frac{0}{0}-2 \mathbf{i}-(2 x-3) \mathbf{j}+(2-2 y) \mathbf{k} $$ \ \hline$$ (-1) \mathbf{i}+(0) \mathbf{j}+(2) \mathbf{k} $$ \ $$ (0) \mathbf{i}-(3) \mathbf{j}+(0) \mathbf{k} $$\end{tabular} \ \hline\end{tabular}

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\begin{tabular}{l} Dado el campo vectorial $$ \mathbf{F}(x, y, z)=\left(y^{2}+3 zight) \mathbf{i}+(2 x+z) \mathbf{j}+\left(x^{2}-yight) \mathbf{k} $$ \ ¿cuál es el rotacional $$ 
abla 	imes F $$ ? \ \begin{tabular}{l}$$ 0-1 \mathbf{i}-(2 x-3) \mathbf{j}+(2-2 y) \mathbf{k} $$ \ $$ \frac{0}{0}-2 \mathbf{i}-(2 x-3) \mathbf{j}+(2-2 y) \mathbf{k} $$ \ \hline$$ (-1) \mathbf{i}+(0) \mathbf{j}+(2) \mathbf{k} $$ \ $$ (0) \mathbf{i}-(3) \mathbf{j}+(0) \mathbf{k} $$\end{tabular} \ \hline\end{tabular}

UpStudy AI · Accepted Answer

Planteamos el campo vectorial  
$$
\mathbf{F}(x,y,z)=(y^2+3z)\mathbf{i}+(2x+z)\mathbf{j}+(x^2-y)\mathbf{k}
$$
y lo escribimos en la forma $$\mathbf{F}=(P,Q,R)$$ con
$$
P=y^2+3z,\quad Q=2x+z,\quad R=x^2-y.
$$

Recordamos que el rotacional se define como
$$
\nabla\times\mathbf{F}=\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\right)\mathbf{i}+\left(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\right)\mathbf{j}+\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\mathbf{k}.
$$

1. Calculemos el componente $$\mathbf{i}$$:
   - $$\displaystyle \frac{\partial R}{\partial y}=\frac{\partial (x^2-y)}{\partial y}=-1.$$
   - $$\displaystyle \frac{\partial Q}{\partial z}=\frac{\partial (2x+z)}{\partial z}=1.$$
   
   Por lo tanto,
   $$
   \text{Componente } \mathbf{i}:\quad -1-1=-2.
   $$

2. Calculemos el componente $$\mathbf{j}$$:
   - $$\displaystyle \frac{\partial P}{\partial z}=\frac{\partial (y^2+3z)}{\partial z}=3.$$
   - $$\displaystyle \frac{\partial R}{\partial x}=\frac{\partial (x^2-y)}{\partial x}=2x.$$
   
   Así,
   $$
   \text{Componente } \mathbf{j}:\quad 3-2x.
   $$

3. Calculemos el componente $$\mathbf{k}$$:
   - $$\displaystyle \frac{\partial Q}{\partial x}=\frac{\partial (2x+z)}{\partial x}=2.$$
   - $$\displaystyle \frac{\partial P}{\partial y}=\frac{\partial (y^2+3z)}{\partial y}=2y.$$
   
   Luego,
   $$
   \text{Componente } \mathbf{k}:\quad 2-2y.
   $$

Reuniendo los resultados, tenemos
$$
\nabla\times \mathbf{F}=-2\,\mathbf{i}+(3-2x)\,\mathbf{j}+(2-2y)\,\mathbf{k}.
$$

Entre las opciones dadas, la opción que presenta los mismos componentes es aquella que indica:
$$
\frac{0}{0}-2\,\mathbf{i}-(2x-3)\,\mathbf{j}+(2-2y)\,\mathbf{k}.
$$
Recordando que $$-(2x-3)= 3-2x$$, ésta es la respuesta correcta.
El rotacional de $$ \mathbf{F} $$ es $$ -2\,\mathbf{i} + (3 - 2x)\,\mathbf{j} + (2 - 2y)\,\mathbf{k} $$.

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