Define, on $\mathbb{R}^3 \setminus \{(0,0,z) \in \mathbb{R}^3 \ \mid \ z \in \mathbb{R} \}$, the vector fields $$G(x,y,z) = \left(\frac{x}{x^2+y^2}, \frac{2y}{x^2+y^2}, 2 \right) $$ and $$H(x,y,z) = \left(\frac{-y}{x^2+2y^2}, \frac{x}{x^2+2y^2},3 \right). $$ How do we compute the line integrals of $G$ and $H$ over the closed curve at the intersection of the following surfaces: $$x+y+z = 2 \text{ and } z = x^2+y^2. $$
What is an easy way to solve these types of questions? We can see that $\text{curl}(G) = \text{curl}(H) = 0$, so an idea could be to use the theorem of Stokes (although, I don't think that $H$ is conservative). However, what would be a good surface to choose? The vector fields are not defined on the entire of $\mathbb{R}^3$, so the circle that is enclosed by the curve is not a good choice. Also, a parametrization of the curve would lead to integrals that are hard to solve.
It's pretty tricky to apply Stokes' theorem here, but it can be done. I will only work with the vector field $H$, since it can be done similarly for $G$. I will also remark that the integral you have to compute does not depend on the curve (it could've been any other simple curve that "goes around" the $z$-axis).
Denote by $K$ the curve from the hypothesis. Let $\Gamma$ be the projection of $K$ on the plane $z = 0$ (or any other horizontal plane), i.e. $$\Gamma = \{(x,y,0) \in \mathbb{R}^3 \ \mid \ (x,y,z) \in K \text{ for some } z \in K \}. $$ Now, let $S$ be the vertical cylindrical surface with boundaries $K$ and $\Gamma$, so $$S = \{(x,y,z) \in \mathbb{R}^3 \ \mid \ (x,y,0) \in \Gamma \text{ and } z \in [0,Z_0], \text{ with } (x,y,Z_0) \in K \}. $$
Let $\overrightarrow{n}$ be the outward pointing unit normal vector on $S$ and let $K$ and $\Gamma$ have the induced orientations. Notice that $H$ is defined everywhere on $S$, so from Stokes' theorem, we have that $$\int_K H ds + \int_\Gamma H ds = \int_S \text{curl}(H) \cdot \overrightarrow{n} d\sigma = 0, $$ so $$\int_K Hds = - \int_\Gamma Hds. $$
So, we only need to compute the integral of $H$ over the curve $\Gamma$, which is still not a small task. However, we can make another trick with Stokes' theorem, namely, we enclose the singularity point of $H$ (i.e. the origin) in a "convenient" surface. Convenient here means something like $x^2 + 2y^2 \leq \varepsilon^2$, for some very small $\varepsilon > 0$.
Let $L$ be the curve $x^2 + 2y^2 = \varepsilon^2, z = 0$, with the reversed orientation of $\Gamma$, and let $U$ be the surface of the annulus with boundaries $L$ and $\Gamma$ (I let you find the concrete definition with formulas for $U$), with the outward pointing unit normal $\overrightarrow{T}$ on $U$ defined such that the orientations of $\Gamma$ and $L$ are induced by $\overrightarrow{T}$. Again, $H$ is defined on $U$, so we can apply Stokes' theorem: $$\int_\Gamma Hds + \int_L Hds = \int_U \text{curl}(H) \cdot \overrightarrow{T} d\sigma = 0, $$ so $$\int_\Gamma Hds = - \int_L Hds, $$ so we finally have that $$\int_K Hds = \int_L Hds. $$
Now, to compute the integral of $H$ over $L$, do it via the standard parametrization $$\text{par}: x = \varepsilon cos(t), y = \frac{\varepsilon}{\sqrt{2}} \sin(t), t \in [0,2\pi), $$ so $$\int_L Hds = \int_0^{2\pi} H(\text{par}(t)) \cdot \frac{d}{dt}(\text{par}(t))dt, $$ which is easy, because we chose $L$ as a "convenient" curve for $H$. Note that the final answer does not depend on the $\varepsilon$.
As a final remark: this method only really works because $\text{curl}(H) = 0$. If this were not the case, you would have to actually compute the surface integrals, which can be difficult, and these entire tricks are specifically meant so that you don't have to compute difficult integrals.