I was reviewing some class notes and I found the following problem: Let $$F(x,y,z)=\left(\frac{-yz}{x^2+y^2},\frac{xz}{x^2+y^2},e^z\right)$$ and let $\gamma:\left\{\begin{array}{rcl} z+1&=&x^2+y^2\\ z&=&3 \end{array}\right.$ a curve traversed once counterclockwise when viewed from the point $(0,0,4)$. Calculate $\displaystyle\int_CF\cdot dr$.
I have used Stokes' theorem arriving at the following: $$\displaystyle\int_CF\cdot dr=\iint_R rot F\cdot \vec{n} dA=\int_R\langle \frac{-x}{x^2+y^2},\frac{-y}{x^2+y^2},0\rangle\cdot \langle 0,0,1\rangle dA=0.$$
But my doubt enters when I see that the vector field F is not continuous on the z axis, and the surface passes through $(0,0,-1)$. So my result is still valid? or how can I solve this exercise?
As the vector field is not defined at any point on z-axis, I can suggest below two approaches.
Using first approach,
$z +1 = x^2 + y^2 \implies r^2 = 4 ~$ at $z = 3$
$ \lambda(t) = (2 \cos t, 2 \sin t, 3); t \in (0, 2\pi)$
$\lambda'(t) = (- 2 \sin t, 2 \cos t, 0)$
$ \displaystyle F(\lambda(t)) = \left(- \frac{6 \sin t}{4}, \frac{6 \cos t}{4}, e^3\right)$
$\displaystyle \int_C \vec F (\lambda(t)) \cdot \lambda'(t) ~ dt = 6 \pi$
Using second approach, we parametrize the cylindrical surface $S$ as,
$\phi(z, t) = (2 \cos t, 2 \sin t, z); t \in (0, 2\pi), z \in (0, 3)$
$\phi_z \times \phi_t = (- 2 \cos t, - 2 \sin t, 0)$
$ \displaystyle \nabla \times \vec F = \left( - \frac{\cos t}{2}, - \frac{\sin t}{2}, 0 \right)$
So the surface integral is,
$ \displaystyle \int_0^3 \int_0^{2\pi} (\nabla \times \vec F) \cdot (\phi_z \times \phi_t) ~ dt ~ dz = 6 \pi$
Please note the cylindrical surface has two boundaries, $C: r = 2, z = 3$ and $C_0: r = 2, z = 0$.
We have, $~\displaystyle \int_C \vec F \cdot dr - \int_{C_0} \vec F \cdot dr = \iint_S (\nabla \times \vec F) \cdot dS = 6 \pi$
You can easily see that the line integral over $C_0$, boundary at $z = 0$, is zero.
That leads to $\displaystyle \int_C \vec F \cdot dr = 6 \pi$