A Möbius strip in $\mathbb{R}^{3}$ is parametrized by

$$\sigma(u, v)=(Q(u, v) \sin u, Q(u, v) \cos u, v \cos (u / 2))$$

for $(u, v) \in U=(0,2 \pi) \times \mathbb{R}$, where $Q \equiv Q(u, v)=2-v \sin (u / 2)$. Show that the Gaussian curvature is

$$K=\frac{-1}{\left(v^{2} / 4+Q^{2}\right)^{2}}$$

at $(u, v) \in U$