Here's an example of what I mean: Let $$\vec{F} = \bigg(\frac{2z}{1+y}+\sin(x^2),\frac{3z}{1+x}+\sin(y^2),5(x+1)(y+2)\bigg)$$ and $C$ be the oriented curve consisting of four line segments from $(0,0,0)$ to $(2,0,0)$, from $(2,0,0)$ to $(0,0,2)$, from $(0,0,2)$ to $(0,3,0)$ and from $(0,3,0)$ to $(0,0,0)$.
How would I implement stokes' theorem to solve for the work integral $\int_C \vec{F}d\vec{r}$? (assuming Stoke's Theorem is applicable here...)
Here's some hint (expect some calculation mistakes).
By Stokes' theorem $$\oint_C\vec{F}\cdot d\vec{r}=\iint_S\big(\nabla\times\vec{F}\big)\cdot\hat{n}dS$$ where $\hat{n}$ is the positive (outward drawn) normal to the surface $S$.
If you trace out the line segments, you would get a surface $S=S_1\cup S_2$ where $S_1$ is a triangle on the $yz$-plane with vertices $(0,0,0)$, $(0,0,2)$ and $(0,3,0)$ and $S_2$ is a triangle on the $xz$-plane with vertices $(0,0,0)$, $(2,0,0)$ and $(0,0,2)$.
Now one can write $\iint_S\big(\nabla\times\vec{F}\big)\cdot\hat{n}dS$ as $\iint_{S_1}\big(\nabla\times\vec{F}\big)\cdot\hat{n}dS_1+\iint_{S_2}\big(\nabla\times\vec{F}\big)\cdot\hat{n}dS_2$ and note that on $S_1$, $\hat{n}=\hat{i}$ and on $S_2$, $\hat{n}=\hat{j}$.
$$\nabla\times\vec{F}=\left(5(x+1)-\frac{3}{1+x}\right)\hat{i}+\left(\frac{2}{1+y}-5(y+2)\right)\hat{j}+\frac{z}{(1+x)^2}\hat{k}.$$
Then the line integral equals \begin{align} \iint_{S_1}\big(\nabla\times\vec{F}\big)\cdot\hat{i}dS_1+\iint_{S_2}\big(\nabla\times\vec{F}\big)\cdot\hat{j}dS_2=\iint_{S_1}\left(5(x+1)-\frac{3}{1+x}\right)dydz+\iint_{S_2}\left(\frac{2}{1+y}-5(y+2)\right)dxdz \end{align} Now we can calculate these two surface integrals as double integrals with appropriate limits of integration.