Question

Calculate curls of the following vector functions (a) AG) 4x3 - 2x2-yy + xz2 2

Answer 1

The curl is
`0 \hat x -z^2 \hat y -4xy \hat z`

Step-by-step explanation:

Given the vector function

`\vec A (\vec r) =4x^3 \hat{x}-2x^2y \hat y+xz^2 \hat z`

We can calculate the curl using the definition

`\\abla * \vec A (\vec r ) = \left|\begin{array}{ccc}\hat x&\hat y&\hat z\\\partial/\partial x&\partial/\partial y&\partial/\partial z\\A_x&X_y&A_z\end{array}\right|`

Thus for the exercise we will have

`\\abla * \vec A (\vec r ) = \left|\begin{array}{ccc}\hat x&\hat y&\hat z\\\partial/\partial x&\partial/\partial y&\partial/\partial z\\4x^3&-2x^2y&xz^2\end{array}\right|`

So we will get

`\\abla  * \vec A (\vec r )= \left( \cfrac{\partial}{\partial y}(xz^2)-\cfrac{\partial}{\partial z}(-2x^2y)\right) \hat x - \left(\cfrac{\partial}{\partial x}(xz^2)-\cfrac{\partial}{\partial z}(4x^3) \right) \hat y + \left(\cfrac{\partial}{\partial x}(-2x^2y)-\cfrac{\partial}{\partial y}(4x^3) \right) \hat z`

Working with the partial derivatives we get the curl

`\\abla  * \vec A (\vec r )=0 \hat x -z^2 \hat y -4xy \hat z`