Deciding a surface in Stokes Theorem

Question

Compute $\int_C (y+z,z+x,x+y) d\vec{r}$, where $C$ is the intersection of the cylinder $x^2 +y^2 = 2y$ and the plane $y = z$.

Is is true that all I can do is apply Stokes Theorem: since $C$ is a closed curve, I need to choose a surface $S$ bounded by $C$ such that orientations of $C$ and $S$ are compatible? Since the curl of $F$ is $0$, then the integral is $0$.

My confusion is how to choose the surface $S$ in this question. Can I choose $S$ to be $\langle r\cos(\theta),1-r\sin(\theta),1-r\sin(\theta) \rangle$, where $0\leq r \leq1$ and $0 \leq \theta \leq 2\pi$?

Answer 1

There are two approaches here to show that the integral is $0$, which are more or less equivalent:

1. The vector field $(y+z, x+z, x+y)$ is the gradient field of $f(x,y,z) := xy + xz + yz$. By the gradient theorem, integrating this vector field along any curve from point $P$ to point $Q$ will give $f(Q)-f(P)$. In particular, a closed curve such as the curve $C$ in the problem is a curve that starts and ends at the same point, and so we get $f(P) - f(P) = 0$.
2. By Stokes' theorem, the line integral of $(y+z, x+z, x+y)$ around $C$ is equal to the integral of the curl of $(y+z, x+z, x+y)$ across any surface whose boundary is $C$. Here, the curl is $0$, and since the vector field and its curl are defined on all of $\mathbb R^3$, it does not really matter which surface with boundary $C$ we pick; we'll get $0$.

The surface parameterized by $\langle r \cos \theta, 1 - r \sin\theta, r\sin\theta\rangle$, as pointed out in the comments, is not quite right: it lies in the plane $y+z=1$, not $y=z$. If we correct it to $\langle r \cos\theta, 1-r\sin\theta, 1-r\sin\theta\rangle$, then this is a perfectly good choice; it is the intersection of the solid cylinder $x^2 + y^2 \le 2y$ with the plane $y=z$. Since we're not going to need to do the hard work of evaluating this integral, finding the parameterization is wasted effort, though; we might as well say, "Let $S$ be the intersection of the solid cylinder $x^2 + y^2 \le 2y$ with the plane $y=z$."

The only reason to even specify a specific surface $S$ is to demonstrate that there's at least one valid choice of $S$. A choice of $S$ is valid if it avoids any problem points for our curl integral, and here there are no such points. So we can be confident that some surface $S$ exists without specifically finding it. (If we use the gradient theorem instead, we don't need a surface at all!)

Finally, though in this case it did not matter, it's worth pointing out that both the line integral and the curve integral are oriented integrals; if we were expecting a nonzero answer, we'd want to make sure the sign is correct by giving $C$ and $S$ compatible orientations.