This is an exercise from my Differential Geometry course:
Define the function $\Phi:\left]0,2\right[\times \left]-\pi/2,\pi/2\right[\longrightarrow \mathbb{R}^2$, $\Phi(\rho,\theta)=(\rho\cos \theta,\rho\sin \theta)$, that covers the right half of the disk $D(0,2)$, and call this half-disk $D$. Prove that the function $$F:D\longrightarrow \mathbb{R}^3,\quad F(\Phi(\rho,\theta))=\left(\frac{1}{2} \rho \cos (2 \theta ),\frac{1}{2} \rho \sin (2 \theta ),\frac{\sqrt{3} }{2}\rho\right),$$ is a local isometry between $D$ and its image $F(D)$. It can be easily seen that $F(D)$ lies in the revolution cone $3(x^2+y^2)=z^2$.
Here's my attempt to solve this problem:
We try to prove that $dF_pv\cdot dF_pv=v\cdot v$, for every $v\in \mathbb{R}^2$. First, we calculate the differential $dF_{p}$ at a point $p=\Phi(\rho,\theta)$: $$dF_p=\left( \begin{array}{cc} \frac{1}{2} \cos (2 \theta ) & -\rho \sin (2 \theta ) \\ \frac{1}{2} \sin (2 \theta ) & \rho \cos (2 \theta ) \\ \frac{\sqrt{3}}{2} & 0 \\ \end{array} \right).$$ Then, we find the value of $dF_pv$, where $v=(x,y)$: $$ dF_pv=\begin{pmatrix}\frac{1}{2}\cos (2 \theta )x-\rho \sin (2 \theta )y \\ \frac{1}{2}\sin (2\theta )x+\rho \cos (2 \theta )y\\\frac{\sqrt{3} }{2}x\end{pmatrix}.$$ Finally, we calculate the product $dF_pv\cdot dF_pv$: $$dF_pv\cdot dF_pv=x^2+\rho^2y^2.$$ However, this product doesn't yield the desired value $v\cdot v=x^2+y^2$. Have I done something wrong? With so many variables maybe I got the concepts messed up.
The metric in polar coordinates is
\begin{bmatrix} 1 & 0 \\ 0 & \rho^2 \end{bmatrix}
For an arbitrary vector $\vec{v}=x\hat{r}+y\hat{\theta}$ the square length of the vector can be calculated as $$ \begin{bmatrix} x & y \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & \rho^2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} $$
Which gives me the answer: $\lvert\lvert\vec{v}\rvert\rvert^2 = x^2+\rho^2y^2$