Specific Examples of Calculations Using the Riemann Metric
Notation
Let’s say we have coordinates for $(u,v)$ as $U$ on a simple surface $\mathbf{x} : U \to \mathbb{R}^{3}$.
$$ \mathbf{x}_{1} := \dfrac{\partial \mathbf{x}}{\partial u}\quad \text{and} \quad \mathbf{x}_{2} := \dfrac{\partial \mathbf{x}}{\partial v} $$
Let’s represent the coefficients of the Riemannian metric as follows.
$$ \begin{align*} g_{ij} =&\ \left\langle \mathbf{x}_{i}, \mathbf{x}_{j} \right\rangle \\ g_{11} =&\ \left\langle \mathbf{x}_{1}, \mathbf{x}_{1} \right\rangle = E \\ g_{12} =&\ g_{21} = \left\langle \mathbf{x}_{1}, \mathbf{x}_{2} \right\rangle = F \\ g_{22} =&\ \left\langle \mathbf{x}_{2}, \mathbf{x}_{2} \right\rangle = G \end{align*} $$
Example
Length of a Curve
Let’s say $U = \left\{ (u,v) : u^{2} + v^{2} \lt 1 \right\}$. Assume a simple surface $\mathbf{x} : U \to \mathbb{R}^{3}$ is given as the following hemisphere.
$$ \mathbf{x}(u,v) = (u, v, \sqrt{1-u^{2}-v^{2}}) $$
Then
$$ \mathbf{x}_{u} = \mathbf{x}_{1} = \left( 1, 0, \dfrac{-u}{\sqrt{1-u^{2}-v^{2}}} \right) \quad \text{and} \quad \mathbf{x}_{v} = \mathbf{x}_{2} = \left( 0, 1, \dfrac{-v}{\sqrt{1-u^{2}-v^{2}}} \right) $$
And let’s say $u, v$ is as follows.
$$ \alpha_{1}(t) = u(t) = t \quad \text{and} \quad \alpha_{2}(t) = v(t) = t^{2} $$
Then, the path in each region according to $0 \lt t \lt \sqrt{\dfrac{\sqrt{5}-1}{2}}$ is shown as follows.
The formula for calculating the length of the blue curve on the hemisphere is as follows.
$$ \int_{a}^{b} \sqrt{ g_{ij} \alpha_{i}^{\prime} \alpha_{j}^{\prime} } dt = \int_{a}^{b} \sqrt{ E(u^{\prime})^{2} + 2F u^{\prime} v^{\prime} + G(v^{\prime})^{2}}dt $$
Calculating each gives the following.
$$ u^{\prime} = 1 \quad \text{and} \quad v^{\prime} = 2t $$
$$ \begin{align*} E =&\ g_{11} = \left\langle \mathbf{x}_{1}, \mathbf{x}_{1} \right\rangle = 1 + \dfrac{u^{2}}{1-u^{2}-v^{2}} = \dfrac{1-v^{2}}{1-u^{2}-v^{2}} = \dfrac{1-t^{4}}{1-t^{2}-t^{4}} \\ F =&\ g_{12} = \left\langle \mathbf{x}_{1}, \mathbf{x}_{2} \right\rangle = \dfrac{uv}{1-u^{2}-v^{2}} = \dfrac{t^{3}}{1-t^{2}-t^{4}} \\ G =&\ g_{22} = \left\langle \mathbf{x}_{2}, \mathbf{x}_{2} \right\rangle = 1 + \dfrac{v^{2}}{1-u^{2}-v^{2}} = \dfrac{1-u^{2}}{1-u^{2}-v^{2}} = \dfrac{1-t^{2}}{1-t^{2}-t^{4}} \end{align*} $$
Therefore, the length of the curve is as follows.
$$ \begin{align*} & \int_{0}^{\sqrt{(\sqrt{5}-1) / 2}} \sqrt{ E(u^{\prime})^{2} + 2F u^{\prime} v^{\prime} + G(v^{\prime})^{2}}dt \\ =&\ \int_{0}^{\sqrt{(\sqrt{5}-1) / 2}} \sqrt{ \dfrac{1-t^{4}}{1-t^{2}-t^{4}}(1)^{2} + 2\dfrac{t^{3}}{1-t^{2}-t^{4}} \cdot 1 \cdot 2t + \dfrac{1-t^{2}}{1-t^{2}-t^{4}}(2t)^{2}}dt \\ =&\ \int_{0}^{\sqrt{(\sqrt{5}-1) / 2}} \sqrt{ \dfrac{-t^{4} + 4t^{2} +1}{1-t^{2}-t^{4}}}dt \end{align*} $$
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Area of a Surface
In the same situation, let’s say $Q= \left\{ (u,v) \in U : 0\le u,v \lt 1 \right\}$. Converting $u, v$ to polar coordinates, since $u = r\cos \theta, v = r\sin \theta$,
$$ \mathbf{x}_{u} = \mathbf{x}_{1} = \left( 1, 0, \dfrac{-u}{\sqrt{1-u^{2}-v^{2}}} \right) = \left( 1, 0, \dfrac{-r\cos\theta}{\sqrt{1-r^{2}}} \right) \\ \mathbf{x}_{v} = \mathbf{x}_{2} = \left( 0, 1, \dfrac{-v}{\sqrt{1-u^{2}-v^{2}}} \right) = \left( 1, 0, \dfrac{-r\sin\theta}{\sqrt{1-r^{2}}} \right) $$
$$ \begin{align*} E =&\ g_{11} = \left\langle \mathbf{x}_{1}, \mathbf{x}_{1} \right\rangle = \dfrac{1-v^{2}}{1-u^{2}-v^{2}} = \dfrac{1-r^{2}\sin^{2}\theta}{1-r^{2}} \\ F =&\ g_{12} = \left\langle \mathbf{x}_{1}, \mathbf{x}_{2} \right\rangle = \dfrac{r^{2}\cos\theta \sin\theta}{1-r^{2}} \\ G =&\ g_{22} = \left\langle \mathbf{x}_{2}, \mathbf{x}_{2} \right\rangle = \dfrac{1-u^{2}}{1-u^{2}-v^{2}} = \dfrac{1-r^{2}\cos^{2}\theta}{1-r^{2}} \end{align*} $$
The formula for calculating the area of region $R = \mathbf{x}(Q)$ on the surface is as follows.
$$ \iint _{Q} \sqrt{g} du dv = \iint _{Q} \left| \mathbf{x}_{1} \times \mathbf{x}_{2} \right| du dv = \iint _{Q} \sqrt{EG-F^{2}} du dv $$
Since $R$ corresponds to the region of $\dfrac{1}{8}$ on the sphere, the integral value should be $\dfrac{4\pi}{8}=\dfrac{\pi}{2}$. When converting to polar coordinates, the Jacobian is $r$, so $dudv=rdrd\theta$ is,
$$ \begin{align*} & \iint _{Q} \sqrt{EG-F^{2}} du dv \\ =&\ \int \limits_{\theta = 0}^{\pi / 2} \int \limits_{r=0}^{1} \sqrt{EG-F^{2}} rdr d\theta \\ =&\ \int \limits_{\theta = 0}^{\pi / 2} \int \limits_{r=0}^{1} \dfrac{\sqrt{(1-r^{2}\cos^{2}\theta)(1-r^{2}\sin^{2}\theta)-r^{4}\cos^{2}\theta \sin^{2}\theta}}{1-r^{2}} rdr d\theta \\ =&\ \int \limits_{\theta = 0}^{\pi / 2} \int \limits_{r=0}^{1} \dfrac{\sqrt{-r^{2}\cos^{2}\theta - r^{2}\sin^{2}\theta + 1}}{1-r^{2}} rdr d\theta \\ =&\ \int _{\theta = 0}^{\pi / 2}d\theta \int _{r=0}^{1} \dfrac{\sqrt{1- r^{2}}}{1-r^{2}} rdr \\ =&\ \dfrac{\pi}{2} \int _{r=0}^{1} \dfrac{r}{\sqrt{1-r^{2}}}dr \\ =&\ \dfrac{\pi}{2} \left[ -\sqrt{1-r^{2}} \right]_{0}^{1} \\ =&\ \dfrac{\pi}{2} \end{align*} $$
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