Does the following integral vanish?

Question

Let $C$ be the meridian determined by the plane $x+y+z = 0$ on the sphere $\mathbb{S} \subset \mathbb{R}^3$ of radius $\rho$ centered at the origin. Let $$\omega = (y+z)\mathrm{d}x + (x+z)\mathrm{d}y + (x+y)\mathrm{d}z$$ Find $\int_C \omega$.

Now what I did was first computing the exterior derivative of $\omega$, and I got that $\mathrm{d}\omega = 0$. Now, I think that closed forms on $\mathbb{R}^3$ are exact (if anyone could point me out the reference to that or if I'm misquoting I'd be grateful). Either way, $$f(x,y,z) = xy +xz +yz + \text{cte}$$ satisfies $\mathrm{d}f = \omega$, so by Stoke's Theorem $$\int_C \omega= \int_{C} \mathrm{d}f = \int_{\partial C}f = \int_{\emptyset} f = 0$$ Is this correct?

Answer 1

You don't even need to argue that $\omega$ is exact in order to see the integral is zero. Consider $\omega$ as one-form on $\mathbb{R}^3$. We can think $C$ as a curve in $\mathbb{R}^3$ and then take the integral of $\omega$ on $C$ in $\mathbb{R}^3$. To show this integral is zero, we can let $D$ be the disc which lies on $x + y + z = 0$ and whose boundary is $C$. Since $d\omega = 0$ on $\mathbb{R}^3$, the form is closed. Note that $\omega$ is defined everywhere and in particular on $D$ and then you apply Stokes' theorem for $D$ and get $$ \int_C \omega = \int_{\partial D} \omega = \int_D d\omega = \int_D 0 = 0. $$