Calculo Diferencial e Integral 3 - Rotacional

Nesse exercício vamos estudar a operação de rotacional.

O rotacional de uma função pode ser calculado da seguinte forma:

$$\nabla\times F=\begin{vmatrix}i&j&k\\\\\dfrac{\partial}{\partial x}&\dfrac{\partial}{\partial y}&\dfrac{\partial}{\partial z}\\\\F_1&F_2&F_3\end{vmatrix}=\left(\dfrac{\partial F_3}{\partial y}-\dfrac{\partial F_2}{\partial z}\right)i+\left(\dfrac{\partial F_1}{\partial z}-\dfrac{\partial F_3}{\partial x}\right)j+\left(\dfrac{\partial F_2}{\partial x}-\dfrac{\partial F_1}{\partial y}\right)k$$

Para o nosso caso:

$$\nabla\times F=\left(\dfrac{\partial}{\partial y}(-x^2y)-\dfrac{\partial}{\partial z}0\right)i+\left(\dfrac{\partial}{\partial z}(xyz)-\dfrac{\partial}{\partial x}(-x^2y)\right)j+\left(\dfrac{\partial}{\partial x}0-\dfrac{\partial}{\partial y}(xyz)\right)k$$

Finalmente:

$$\boxed{\nabla\times F=-x^2i+3xyj-xzk}$$

Nesse exercício vamos estudar a operação de rotacional.

O rotacional de uma função pode ser calculado da seguinte forma:

$$\nabla\times F=\begin{vmatrix}i&j&k\\\\\dfrac{\partial}{\partial x}&\dfrac{\partial}{\partial y}&\dfrac{\partial}{\partial z}\\\\F_1&F_2&F_3\end{vmatrix}=\left(\dfrac{\partial F_3}{\partial y}-\dfrac{\partial F_2}{\partial z}\right)i+\left(\dfrac{\partial F_1}{\partial z}-\dfrac{\partial F_3}{\partial x}\right)j+\left(\dfrac{\partial F_2}{\partial x}-\dfrac{\partial F_1}{\partial y}\right)k$$

Para o nosso caso:

$$\nabla\times F=\left(\dfrac{\partial}{\partial y}(-x^2y)-\dfrac{\partial}{\partial z}0\right)i+\left(\dfrac{\partial}{\partial z}(xyz)-\dfrac{\partial}{\partial x}(-x^2y)\right)j+\left(\dfrac{\partial}{\partial x}0-\dfrac{\partial}{\partial y}(xyz)\right)k$$

Finalmente:

$$\boxed{\nabla\times F=-x^2i+3xyj-xzk}$$

Rotacional  F =    (dh/dy)-(dg/dz)i+(df/dz-dh/dx)j+(dg/dx-df/dy)k

                      (-x² - 0)i + (xy - (-2xy)j + (0-xz)   =>     rotacional   F =    -x² i  +  3xy j  - xz K