[size=150]Si [math]F=F_1i+F_2j+F_3k[/math], el [b]rotacional[/b] de [b]F[/b] es el vector campo[br][br][math]rot\left(F\right)=\bigtriangledown\times F=\left(\frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z}\right)i+\left(\frac{\partial F_1}{\partial z}-\frac{\partial F_3}{\partial x}\right)j+\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)k.[/math][/size]

[size=150][b]13.-[/b][/size][math]F\left(x,y,z\right)=xi+yj+zk[/math][br][br]Utilizando la definición, tenemos lo siguiente:[br][br] [math]rot\left(F\right)=\left(\frac{\partial\left(z\right)}{\partial y}-\frac{\partial\left(y\right)}{\partial z}\right)i+\left(\frac{\partial\left(x\right)}{\partial z}-\frac{\partial\left(z\right)}{\partial x}\right)j+\left(\frac{\partial\left(y\right)}{\partial x}-\frac{\partial\left(x\right)}{\partial y}\right)k.[/math][br][br]Resolviendo las derivadas:[br][br] [math]rot\left(F\right)=\left(0-0\right)i+\left(0-0\right)j+\left(0-0\right)k.[/math][br][br]Por lo tanto, el resultado del rotacional de[math]F\left(x,y,z\right)[/math] es:[br][br] [math]rot\left(F\right)=0i+0j+0k=\left(0,0,0\right)[/math]

[size=150][b]15.-[/b][/size][math]F\left(x,y,z\right)=\left(x^2+y^2+z^2\right)\left(3i+4j+5k\right)[/math][br][br]Simplificamos:[br][br] [math]F\left(x,y,z\right)=3\left(x^2+y^2+z^2\right)i+4\left(x^2+y^2+z^2\right)j+5\left(x^2+y^2+z^2\right)k[/math][br][br]Utilizando la definición, tenemos lo siguiente:[br][br][math]rot\left(F\right)=\left(\frac{\partial\left(5\left(x^2+y^2+z^2\right)\right)}{\partial y}-\frac{\partial\left(4\left(x^2+y^2+z^2\right)\right)}{\partial z}\right)i+\left(\frac{\partial\left(3\left(x^2+y^2+z^2\right)\right)}{\partial z}-\frac{\partial\left(5\left(x^2+y^2+z^2\right)\right)}{\partial x}\right)j+\left(\frac{\partial\left(4\left(x^2+y^2+z^2\right)\right)}{\partial x}-\frac{\partial\left(3\left(x^2+y^2+z^2\right)\right)}{\partial y}\right)k.[/math][br][br]Resolviendo las derivadas:[br][br] [math]rot\left(F\right)=\left(10y-8z\right)i+\left(6z-10x\right)j+\left(8z-6y\right)k[/math][br][br]Por lo tanto, el rotacional de [math]F\left(x,y,z\right)[/math] es:[br][br] [math]rot\left(F\right)=\left(10y-8z\right)i+\left(6z-10x\right)j+\left(8z-6y\right)k[/math]