Parametrising a given surface to verify Stokes' Theorem 
Question
  Let $M := \{ (x,y,z) \in \mathbb{R}^{3}: 4z^{2} - x^{2} = 1 \text{ and } x^{2} + y^{2} <1 \text{ and } z > 0 \}$ be a surface in $\mathbb{R}^{3}$. Verify Stokes' Theorem for $\mathbf{F} = (xy, -2z^{2}, 1)$ on $M$.

I am having trouble finding the correct surface parametrisation for $M$, because of the $4z^{2} - x^{2} = 1$ condition I tried $z = \frac{1}{2} \cosh(t)$ and $x = \sinh(t)$ but I cannot figure out then how to find the values that $t$ must take or to incorporate the unit circle constraint.
How can I work this out?
 A: Since you have the condition $z>0$, you can use $x$ and $y$ as parameters and parametrize the surface $M$ as follows:
\begin{cases}
x=x \\
y=y \\
z= \sqrt{\frac{1+x^2}{4}}
\end{cases}
with $(x,y)\mid x^2+y^2 \le 1$.
For any curve $C$ that bounds this surface, e.g., the circle $x^2+y^2 =1$, you have to verify that the following holds
$$
\oint_C \vec{F}\cdot d\vec{r} = \iint_M \nabla \times\vec{F}\cdot d\vec{S}
$$
For the left hand term, since you are on a unit circle at $z=0$, compute
$$
\oint_C \vec{F}\cdot d\vec{r} = \oint_0^{2\pi} \pmatrix{\cos t \sin t \\ 0 \\ 1}\cdot \pmatrix{- \sin t \\ \cos t \\0}dt=0
$$
For the right hand term, use the above parametrization:
\begin{align}
\iint_M \nabla \times\vec{F}\cdot d\vec{S} &= \iint_{x^2+y^2\le 1}\pmatrix{4\sqrt{\frac{1+x^2}{4}} \\0 \\-x}\cdot \pmatrix{1\\0\\\frac{x}{2\sqrt{1+x^2}}} \times \pmatrix{0\\1\\0} dxdy 
\\&= \iint_{x^2+y^2\le 1}\pmatrix{4\sqrt{\frac{1+x^2}{4}} \\0 \\-x}\cdot
\pmatrix{-\frac{x}{2\sqrt{1+x^2}}\\0\\1}dx dy
\\&= \iint_{x^2+y^2\le 1}-x \;dx dy=0
\end{align}
