Stoke's Thm. $\vec F = \langle x^2y,\frac{x^3}{3},xy \rangle $ and $C$ is the boundary of the intersection of $x^2+y^2=1$ and $z = y^2-x^2$

Question

I am having trouble setting up the limit of integration for the following problem.

$\vec F = \langle x^2y,\frac{x^3}{3},xy \rangle $ and $C$ is the boundary of the intersection of $x^2+y^2=1$ and $z = y^2-x^2$.

I found

$$ \operatorname{curl} \vec F = \langle x,-y,0 \rangle $$

and letting $x=\cos \theta, y = \sin \theta , z = z$

the parameterized surface will have a normal vector

$$\vec n = \langle \cos \theta, \sin \theta, 0\rangle$$.

So, what I need to integrate according to Stoke's Thm. is

$$\iint_S \cos^2 \theta\,dS$$

My question here is, I know that $\theta \in [0, 2\pi]$ but what about $z$?

I want to say that the maximum and minimum values $z$ can take is $1$ and $-1$, respectively, so I want to use $\int_{-1}^1$ but I know that is not right.

I also considered the parameterization $\langle\cos \theta, r\sin \theta, -r^2(\cos 2\theta)\rangle$ but I am not confident going that route.

I believe I have a gap in understanding in this portion of "parameterization" of the curve or surface.

May I have some assistance?

Answer 1

Point 1:

This is a line integral over the boundary of a surface (made by intersection with another surface). So if you want to do line integral for the given vector field then parametrize as

$r(\theta) = (\cos\theta, \sin\theta, - \cos2\theta) \ $, as ($z = y^2 - x^2 = - \cos 2\theta$)

But if you are converting this into a double integral using the curl of the vector field, then you have to parametrize as

$r(\rho, \theta) = (\rho \cos\theta, \rho \sin \theta, - \rho^2 \cos 2\theta)$

Point 2:

you cannot assume normal vector to be $(\cos\theta, \sin\theta, 0)$ as it is intersection of two surfaces. Instead, take derivative with respect to $\rho$ and $\theta$ and do a cross product. That is the standard approach.

$r'_{\rho} = (\cos\theta, \sin\theta, - 2\rho\cos2\theta)$
$r'_{\theta} = (-\rho\sin\theta, \rho\cos\theta, 2\rho^2\sin2\theta)$

$r'_{\rho} \times r'_{\theta} = (2\rho^2 \cos\theta, - 2\rho^2\sin\theta, \rho)$

$curl \vec F = (\rho \cos\theta, - \rho \sin\theta, 0)$

So the double integral is,

$\displaystyle \int_0^{2\pi} \int_0^{1} (\rho \cos\theta, - \rho \sin\theta, 0) \cdot (2\rho^2 \cos\theta, - 2\rho^2\sin\theta, \rho) \ d\rho \ d\theta$

$ = \displaystyle \int_0^{2\pi} \int_0^{1} 2 \rho^3 \ d\rho \ d\theta$