Explanation:
Given:
![\textbf{F} = (2xy + z^3)\hat{\textbf{i}} + x^3\hat{\textbf{j}} + 3xz^2\hat{\textbf{k}}](https://img.qammunity.org/qa-images/2022/formulas/mathematics/college/qvi32dhndlsvw8f326u881.png)
This field will have a scalar potential
if it satisfies the condition
. While the first x- and y- components of
are satisfied, the z-component doesn't.
![(\\abla * \textbf{F})_z = \left((\partial F_y)/(\partial x) - (\partial F_x)/(\partial y) \right)](https://img.qammunity.org/qa-images/2022/formulas/mathematics/college/sckigx8lf74gcidyzmrln4.png)
![\:\:\:\:\:\:\:\:\: = 3x^2 - 2x \\e 0](https://img.qammunity.org/qa-images/2022/formulas/mathematics/college/tkp52es6dno96d0ubauuu8.png)
Therefore the field is nonconservative so it has no scalar potential. We can still calculate the work done by defining the position vector
as
![\vec{\textbf{r}} = x \hat{\textbf{i}} + y \hat{\textbf{j}} + z \hat{\textbf{k}}](https://img.qammunity.org/qa-images/2022/formulas/mathematics/college/srukss9a5oph2mr3owqy5l.png)
and its differential is
![\textbf{d} \vec{\textbf{r}} = dx \hat{\textbf{i}} + dy \hat{\textbf{j}} + dz \hat{\textbf{k}}](https://img.qammunity.org/qa-images/2022/formulas/mathematics/college/zxru78v62tbu0oi11y2vio.png)
The work done then is given by
![\displaystyle \oint_c \vec{\textbf{F}} • \textbf{d} \vec{\textbf{r}} = \int ((2xy + z^3)\hat{\textbf{i}} + x^3\hat{\textbf{j}} + 3xz^2\hat{\textbf{k}}) • (dx \hat{\textbf{i}} + dy \hat{\textbf{j}} + dz \hat{\textbf{k}})](https://img.qammunity.org/qa-images/2022/formulas/mathematics/college/lwm8baa5isbw3ns2vpfl6d.png)
![\displaystyle = (x^2y + xz^3) + x^3y + xz^3|_((1, -2, 1))^((3, 1, 4))](https://img.qammunity.org/qa-images/2022/formulas/mathematics/college/4pv4wyxvy69s8gbc7dyehm.png)
![= 422](https://img.qammunity.org/qa-images/2022/formulas/mathematics/college/txa49m8oinnzoh11csbmry.png)