Using Stokes theorem to find the integral of a vector field over the curve of intersection of two surfaces

Question

Find $\int_C{ \vec{F} \cdot \vec{dr}},$ where $F(x, y, z) = \langle 2 x^2 y , 2 x^3 /3, 2xy\rangle$ and $C$ is the curve of intersection of the hyperbolic paraboloid $z = y^2 - x^2$ and the cylinder $x^2 + y^2 = 1$, oriented counterclockwise as viewed from above.

I tried to parametrize the region but have no idea what to do after that.

Answer 1

In the comments we reduced this to integrating $\langle 2x, -2y,0\rangle$ over the part of paraboloid $z=y^2−x^2$ that lies within the cylinder $x^2+y^2=1$. Note that the Stokes theorem was already used, the rest is just a run-of-the-mill computation of a surface integral. So if you are uncertain about how this is done, that would be the topic to revisit. Some points to keep in mind

1. You need a parametric equation $\vec r(u,v)$ and the parameter domain in the $uv$-plane. So there will be two distinct steps: first you'll get a parametric equation, and then figure out the values the parameters take.
2. The formula for $\vec r(u,v)$ comes from the surface over which you integrate. Do not combine the equations of both surfaces just because you have them. Use only the one over which you will integrate, which is the paraboloid.
3. The parameter domain is where you bring the other surface into consideration. Think of it as a cookie-cutter slicing the first surface. How does it cut through the first one? Draw a sketch.

Answer 2

Since the intersection lies on the surface $z = y^2 - x^2$ and the dependence of $z$ is linear, this means once you know how to parametrize $(x,y)$, you know how to parametrize $z$.

Since the other surface is $x^2 + y^2 = 1$ which doesn't depend on $z$ explicitly, we can parametrize it like what we do to a circle in the plane.

The question tells us when you look at it from the top, the intersection runs counterclockwisely, this means the ordinary parametrization of unit circle in the plane $$[0,2\pi] \ni \theta \quad\mapsto\quad (x,y) = (\cos\theta, \sin\theta) \in \mathbb{R}^2$$

provide the parametrization we need for the $(x,y)$ part.

When $(x,y) = ( \cos\theta,\sin\theta )$, $z = y^2 - x^2 = -\sin^2\theta + \cos^2\theta = -\cos(2\theta)$. The parametrization we need for the intersection is given by $$[0,2\pi] \ni \theta \quad\mapsto\quad (x,y,z) = (\cos\theta, \sin\theta, -\cos(2\theta))\in \mathbb{R}^3$$

and the line integral becomes

$$\int_C \vec{F}(\vec{r})\cdot \vec{r} = \int_0^{2\pi} \left( 2 x^2 y \frac{dx}{d\theta} + \frac{2}{3} x^3 \frac{dy}{d\theta} + 2xy \frac{dz}{d\theta}\right) d\theta$$ Before we rush to substitute above parametrization into RHS and blindly integrate. One should sit back and analysis the integrand first. We find

$$2 x^2 y \frac{dx}{d\theta} + \frac{2}{3} x^3 \frac{dy}{d\theta} = \frac{d}{d\theta}\left( \frac23 x^3 y \right)$$

is a total differential and its integral over any closed contour vanishes. As a result, your integral reduces to

$$\int_0^{2\pi} 2\sin\theta\cos\theta d(-\cos(2\theta)) = 2\int_0^{2\pi} \sin^2(2\theta) d\theta = 2\pi$$