Use Stokes' theorem to calculate the surface integral
$$\iint_D\text{curl}\,\mathbf F \cdot\mathrm d\mathbf S$$ where $\mathbf F(x, y, z) = (xz,xe^z,-y)$ and $S = \{(x,y,z):z=-1+x^2+y^2\ \text{and}\ z\leq0\}$, with positive orientation.
Since $z=0$, then $x^2+y^2=1$ and
$$\mathbf r(t) = (\cos(t),\sin(t),0) \\ \mathbf r'(t)=(-\sin(t),\cos(t),0)$$
So:
$$\begin{align}\iint_D\text{curl}\,\mathbf F\cdot\mathrm d\mathbf S &= \int_0^{2\pi} \mathbf F(\mathbf r(t))\cdot \mathbf r'(t)\,\mathrm dt \\ &= \int_0^{2\pi}\cos^2t\,\mathrm dt = \pi\end{align}$$
Could someone check if this is right?