We will take exercises from Boas 10.8. Try to find the $ds^2$ (same as the metric tensor). The scale factor $H_i$ (not to be confused with my symbol for the dual basis), the volume or the area element, and the $\bf a$ factor.

#6, #8, #10, #15

#6: The new coordinate $(u,v,z)$ is related to the Cartesian one by $x = (1/2)(u^2-v^2),\quad y = uv, \quad z = z$

Let's compute $ds^2$. I will compute it in two ways, the Boas way, and the math way, you may compare.

In Boas way, we write $dx = u du - v dv, \quad dy = u dv + v du \quad dz = dz$ and we plug in $ds^2 = dx^2 + dy^2 + dz^2$, to get (do the expansion, witness some cancellation) $ds^2 = (u du - v dv)^2 + (u dv + v du)^2 + dz^2 = (u^2+v^2) du^2 + (u^2+v^2) dv^2 + dz^2$

In the math way, we need to compute the coordinate vector fields $\d_u, \d_v, \d_z$ in terms of $\d_x, \d_y, \d_z$ $\begin{cases} \d_u = \d_u(x) \d_x + \d_u(y) \d_y + \d_u(z) \d_z = u \d_x + v \d_y \cr \d_v = \d_v(x) \d_x + \d_v(y) \d_y + \d_v(z) \d_z = - v\d_x + u \d_y \cr \d_z = \d_z \end{cases}$ Thus, we have $g(\d_u, \d_u) = u^2 + v^2, \quad g(\d_v, \d_v) = u^2 + v^2, \quad g(\d_u, \d_v) = 0, \quad g(\d_z, \d_z)=1$ Hence, we can write the tensor $g$ as $g = (u^2+v^2) du \otimes du + (u^2+v^2) dv \otimes dv + dz \otimes dz.$

You may compare $ds^2$ and $g$, and notice that, they are talking about the same thing, with slightly different notations.

Note that we have orthogonal curvilienar coordinate, that means $ds^2$ has no 'mixed term', like $du\,dv, du\,dz, dv\,dz$, or equivalently, $\d_u, \d_v, \d_z$ are perpendicular to each other, $g(\d_u, \d_v) = 0$ etc.

The scale factors $H_u = \| \d_u \| = \sqrt{ g(\d_u, \d_u) } = \sqrt{u^2 + v^2}$ similarly $H_v = \sqrt{u^2+v^2}, H_z = 1$.

The volume form for orthogonal coordinates are easy, $d Vol_g = H_u H_v H_z du dv dz = (u^2+v^2) du dv dz$ Or, we can use $d Vol_g = \sqrt{\det g} du dv dz = \sqrt{(u^2+v^2) (u^2+v^2) } dudvdz = (u^2+v^2) du dv dz.$