I want to use Stokes's Theorem to find the circulation of the vector field $ (2yz,\cos(y),\sin(z))$ around the curve $ C$ given by $ z=3x^2+y^2$ and $ z=4-x^2-3y^2$. Equating $ z=3x^2+y^2$ with $ z=4-x^2-3y^2$ I obtain the equation of the cylinder $x^2+y^2=1$ .
I know stokes's theorem is given as the following:
$$ \iint \nabla \times F(\vec{X}(s,t))\cdot \vec{n} \, ds \, dt$$
What's the parametrization, $ \vec{X}(s,t)$, of the curve $C$?
Is $C$ parametrized by $ \vec{X}(s,t) = s\cos[t]\hat{i}+ s\sin[t]\hat{j} +(3s^2\cos[t]^2+s^2 \sin[t]^2) \hat{k} $ Where $s$ is the radius $\in [0,1]$ from the equation of cylinder and $t \in [0,2\pi]$ ?
Or is $ C$ parametrized by
$$ \vec{X}(s,t) = s\cos[t]\hat{i}+ s\sin[t]\hat{j} +(4-(s^2 \cos[t]^2) - 3 (s^2 \sin[t]^2) \hat{k} \text{ ?} $$
I highly think that $ \vec{X}(s,t) = s\cos[t]\hat{i}+ s\sin[t]\hat{j} +(4-(s^2 \cos[t]^2) - 3(s^2\sin[t]^2) \hat{k} $ is the parametrization of the curve $ C$.
Is there a correct parametrization? The surface is oriented upward.
$\nabla \times F = (0,y,-z)$
$dS = (-\frac {\partial z}{\partial x}, -\frac {\partial z}{\partial y}, 1)$
And you can use either surface. Any surface with the same contour should give the same result.
$\iint (0,y, -3x^2 -y^2)\cdot (-6x, -2y, 1) dA = \iint (0,y, -4+x^2 - 3y^2)\cdot (2x, 6y, 1) dA\\ \iint -3y^2 -3x^2 \ dA = \iint 9y^2+x^2 -4 \ dA $
The one on the left looks easier. Convert to cylindrical
$\int_0^{2\pi} \int_0^1 -3r^3 \ dr\ dt = -\frac {3\pi}{2}$
And that should equal
$\iint 4 - 9y^2-x^2 \ dA\\ \int_0^1 \int_0^{2\pi} 4r - 9r^3\sin^2 - r^3\cos^2\theta d\theta\ dr\\ \int_0^1 8\pi r - 10\pi r^3 \ dr\\ -\frac {3\pi}{2}$
and that should also equal
$\oint F\cdot dr$
$r = (\cos t, \sin t, 3\cos^2 t + \sin^2 t) = (\cos t, \sin t, 4 - \cos^2 t - 3\sin^2 t) = (\cos t, \sin t, \cos 2t + 2)$
$F = (yz,0,0) + (0,\cos y, \sin z)$ but $(0,\cos y, \sin z) = \nabla (\sin y -\cos z)$ is conservative and can be dropped.
$\int_0^{2\pi} (\sin t)(3 - 2\sin 2t)(-\sin t) \ dt = -\frac {3\pi}{2}$