I have a four-dimensional integration problem (the variables of integration being $z_{14},y_{14},y_{13},z_{13}$) defined over \begin{equation} u>1\land -\frac{1}{u}<z_{14}<\frac{1}{u}\land -\frac{\sqrt{1-u^2 z_{14}^2}}{u}<y_{14}<\frac{\sqrt{1-u^2 z_{14}^2}}{u}\land -\sqrt{1-u^2 \left(y_{14}^2+z_{14}^2\right)}<y_{13}<\sqrt{1-u^2 \left(y_{14}^2+z_{14}^2\right)}\land -\sqrt{u^2 \left(-\left(y_{14}^2+z_{14}^2\right)\right)-y_{13}^2+1}<z_{13}<\sqrt{u^2 \left(-\left(y_{14}^2+z_{14}^2\right)\right)-y_{13}^2+1}, \end{equation} where $u$ is a fifth variable, not to be integrated over.
If I perform the change-of-variables (to two sets of polar coordinates) \begin{equation} \left\{z_{13}\to r_{13} \cos \left(t_{13}\right),z_{14}\to r_{14} \cos \left(t_{14}\right),y_{13}\to r_{13} \sin \left(t_{13}\right),y_{14}\to r_{14} \sin \left(t_{14}\right)\right\}, \end{equation} what should be the new limits of integration over $r_{13}, r_{14}, t_{13}$ and $t_{14}$. (The jacobian for the change-of-variables would, then, be $r_{13} r_{14}$.)
I converted the constraints--given above in Cartesian form--using the two sets of polar coordinates. Then, I re-expressed the constraints using the squares of all the quantities involved, having noted that all the constraints were of the form $-b<a<b$. That is, I transformed the constraints to the form $a^2<b^2$. They, then, drastically simplified--using trigonometric identities--to $r_{13}^2 +r_{14}^2 u^2 <1$, with $u>1$. With this simplification, I was able to immediately answer the companion four-dimensional integration questions put in my subsequent posting Solve a constrained 4D integration problem using either Cartesian or (paired) polar coordinates