Question

determine o rotacional e o divergente de F (x,y,z) = (x²-y)i 4zj x²k

Answer 1

$\mathbf{F}(x,\;y,\;z)=(x^{2}-y)\hat{\mathbf{i}}+(4z)\hat{\mathbf{j}}+(x^{2})\hat{\mathbf{k}}\\ \\ \mathbf{F}(x,\;y,\;z)=(x^{2}-y;\,4z;\,x^{2})$


E temos o operador nabla:

$\nabla=\frac{\partial}{\partial x}\hat{\mathbf{i}}+\frac{\partial}{\partial y}\hat{\mathbf{j}}+\frac{\partial}{\partial z}\hat{\mathbf{k}}\\ \\ \nabla=\left(\frac{\partial}{\partial x};\,\frac{\partial}{\partial y};\,\frac{\partial}{\partial z} \right )$


$\bullet\;\;$ O rotacional é o produto vetorial entre o operador $\nabla$ e o campo vetorial $\mathbf{F}.$ O resultado é um campo vetorial:

$\mathrm{rot\,}\mathbf{F}=\nabla \times \mathbf{F}\\ \\ \mathrm{rot\,}\mathbf{F}=\left(\frac{\partial}{\partial x};\,\frac{\partial}{\partial y};\,\frac{\partial}{\partial z} \right ) \times (x^{2}-y;\,4z;\,x^{2})\\ \\ \mathrm{rot\,}\mathbf{F}=\det\left[\begin{array}{ccc} \hat{\mathbf{i}}&\hat{\mathbf{j}}&\hat{\mathbf{k}}\\ \frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\ x^{2}-y&4z&x^{2} \end{array} \right ]\\ \\ \\ \mathrm{rot\,}\mathbf{F}=\left[\frac{\partial}{\partial y}(x^{2})-\frac{\partial}{\partial z}(4z) \right ]\hat{\mathbf{i}}+\left[\frac{\partial}{\partial z}(x^{2}-y)-\frac{\partial}{\partial x}(x^{2}) \right ]\hat{\mathbf{j}}+\left[\frac{\partial}{\partial x}(4z)-\frac{\partial}{\partial y}(x^{2}-y) \right ]\hat{\mathbf{k}}$


$\mathrm{rot\,}\mathbf{F}=\left[0-4 \right ]\hat{\mathbf{i}}+\left[0-2x \right ]\hat{\mathbf{j}}+\left[0-(-1) \right ]\hat{\mathbf{k}}\\ \\ \mathrm{rot\,}\mathbf{F}=(-4)\hat{\mathbf{i}}+(-2x)\hat{\mathbf{j}}+(1)\hat{\mathbf{k}}\\ \\ \boxed{\begin{array}{c}\mathrm{rot\,}\mathbf{F}=-4\hat{\mathbf{i}}-2x\hat{\mathbf{j}}+\hat{\mathbf{k}} \end{array}}$


$\bullet\;\;$ O divergente é o produto escalar entre o operador $\nabla$ e o campo vetorial $\mathbf{F}.$ O resultado é um campo escalar:

$\mathrm{div\,}\mathbf{F}=\nabla \cdot \mathbf{F}\\ \\ \mathrm{div\,}\mathbf{F}=\left(\frac{\partial}{\partial x};\,\frac{\partial}{\partial y};\,\frac{\partial}{\partial z} \right ) \cdot (x^{2}-y;\,4z;\,x^{2})\\ \\ \mathrm{div\,}\mathbf{F}=\frac{\partial}{\partial x}(x^{2}-y)+\frac{\partial}{\partial y}(4z)+\frac{\partial}{\partial z}(x^{2})\\ \\ \mathrm{div\,}\mathbf{F}=2x+0+0\\ \\ \boxed{\begin{array}{c}\mathrm{div\,}\mathbf{F}=2x \end{array}}$