<s>[INST] Line integral of vector field when Stokes' theorem cannot be applied directly 
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.
 [/INST] If you don't want to use Stoke's theorem then integrate around the path itself.
The cylinder x+ y+ z= 2 and the paraboloid $z= x^2+ y^2$ intersect where $x+ y+ (x^2+ y^2)= 2$.  Complete the squares in both variables: $x^2+ x+ \frac{1}{4}+ y^2+ y+ \frac{1}{4}= 2+ \frac{1}{4}+ \frac{1}{4}$, $(x+ \frac{1}{2})^2+ (y+ \frac{1}{2})^2=\frac{5}{2}$.  That is a circle with center at (1/2, 1/2, z) and radius $\sqrt{\frac{5}{2}}$.  Since $z= 2- x- y$, we can use parametric equations $x= \frac{1}{2}+ \sqrt{\frac{5}{2}}cos(\theta)$, $y= \frac{1}{2}+ \sqrt{\frac{5}{2}}sin(\theta)$, $z= 1- \sqrt{\frac{5}{2}}cos(\theta)- \sqrt{\frac{5}{2}}sin(\theta)$.
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