A special case of the generalized Calabi ansatz is when $L$ is trivial, and $L_{i}=d_{i}L_{0}$ for some positive line bundle $L_{0}$ and positive integers $d_{i}>0$. In this case, we can take $h_{L}=1$, and $h_{L_{i}}$ is the suitable tensor power of $h_{L_{0}}$, where $h_{L_{0}}$ can be chosen to correspond to the Calabi-Yau metric in the class $c_{1}(L_{0})$. The ansatz metric is simply

$$\begin{split}dd^{c}u=\sum\frac{\partial^{2}u}{\partial x_{i}\partial x_{j}}d\log r_{i}\wedge d^{c}\log r_{j}+\sum\frac{\partial u}{\partial x_{i}}dd^{c}\phi_{i}\\ =\sum\frac{\partial^{2}u}{\partial x_{i}\partial x_{j}}d\log r_{i}\wedge d^{c}\log r_{j}+(\sum\frac{\partial u}{\partial x_{i}}d_{i})dd^{c}\phi_{0}.\end{split}$$

Observe that for rank reasons

$$(\sum\frac{\partial^{2}u}{\partial x_{i}\partial x_{j}}d\log r_{i}\wedge d^{c}\log r_{j})^{m+1}=0,\quad(dd^{c}\phi_{0})^{n-m+1}=0.$$

We compute the volume form using binomial expansion

$$(dd^{c}u)^{n}=\frac{n!}{(n-m)!}(\sum\frac{\partial u}{\partial x_{i}}d_{i})^{n-m}\det(D^{2}u)(\prod_{1}^{m}d\log r_{i}\wedge d^{c}\log r_{i})\wedge(dd^{c}\phi_{0})^{n-m}.$$

Since $(dd^{c}\phi_{0})^{n-m}$ already exhaust all base terms, we can replace $\log r_{i}$ by $\log|\xi_{i}|$, and obtain

$$(dd^{c}u)^{n}=\frac{n!}{(n-m)!}(\sum\frac{\partial u}{\partial x_{i}}d_{i})^{n-m}\det(D^{2}u)(\prod_{1}^{m}\frac{\sqrt{-1}}{4\pi}d\log\xi_{i}\wedge d\overline{\log\xi_{i}})\wedge(dd^{c}\phi_{0})^{n-m}.$$

The Calabi-Yau condition on $dd^{c}\phi_{0}$ and the normalization on $\Omega_{Y}$ imply

$$(dd^{c}\phi_{0})^{n-m}=(\int_{Y}c_{1}(L_{0})^{n-m})\sqrt{-1}^{(n-m)^{2}}\Omega_{Y}\wedge\overline{\Omega}_{Y}.$$

The conclusion is that