I am new to vector calculus and found this problem in the textbook which I am not sure how to work with, but I want to learn to do so:
Show details to find the work done by the force field $\mathbf{F} =x^3\,\mathbf{i}+y^3\,\mathbf{j}$ in moving an object from $P(1, 0)$ to $Q(2, 2).$
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Just want to complement @JacobBach's answer, the work done for the force ${\bf F}$ on the path $\gamma$ is calculated as
$$ W = \int_\gamma {\rm d}{\bf x}\cdot{\bf F} \tag{1} $$
If ${\bf F}$ can be written as as the gradient of a potential field
$$ {\bf F} = -\nabla \phi $$
then (1) becomes
$$ W = \int_\gamma {\rm d}{\bf x}\cdot{\bf F} = -\int_\gamma {\rm d}{\bf x}\cdot \nabla\phi = -\int_{\bf a}^{\bf b}{\rm d}\phi= \phi({\bf a}) - \phi({\bf b}) \tag{2} $$
In this case, you can write your force as a gradient of a potential $\phi$,
$\mathbf{F}=-\nabla\phi$,
where $\phi$ is your potential,
$\phi=-\frac{1}{4} (x^4+y^4)$.
You can check that this gives you the right force since $F_i=-\frac{d\phi}{d x}=x^3$ and $F_j=-\frac{d\phi}{d y}=y^3$.
As soon as you have a potential, the work $W_{\text{done by force}}$ done by the force on any path from $P$ to $Q$ is simply the difference in the potential,
$W_{\text{done by force}}=\phi_{start}-\phi_{end}=\phi(P)-\phi(Q) =(-\frac{1}{4}(1^4+0^4))-(-\frac{1}{4}(2^4+2^4))=\frac{31}{4}$.