Find the relationship between $x$ and $y$ so that $$y:=0\rightarrow\frac{\pi}{2}\Leftrightarrow x:=y\rightarrow\frac{\pi}{2}$$
I'm having trouble solving the multivariable calculus if I change the order between ${\rm d}x, {\rm d}y,$ so I try a new approach like finding function $y=\!f_{1}\left ( x \right )$ and $x=\!f_{2}\left ( y \right )$ but unsuccessfully . I need to the help
I can give an example of an integral such that changing order of the integration changes limits like you've said in the title of your question. Consider the following double integral: \begin{equation} \iint\limits_{0<x<y<\frac\pi2}\mathrm dx\mathrm dy \end{equation} That, which I've written below the integral sign is the region of integration, and this integration will find out area of the region given by \begin{equation} A = \left\{(x,y)\Bigg|0<x<y<\frac\pi2\right\} \end{equation} in the plane.
Separating the limits, and doing integration in this order, we get the required integral to be \begin{align} I&=\iint\limits_A\mathrm dx\mathrm dy\\ &=\int_0^\frac\pi2\int_0^y\mathrm dx\mathrm dy\\ &=\int_0^\frac\pi2y\mathrm dy\\ &=\dfrac{\pi^2}{8} \end{align}
Changing the order of integration, we get, \begin{align} I&=\iint\limits_A\mathrm dy\mathrm dx\\ &=\int_0^\frac\pi2\int_x^\frac\pi2\mathrm dy\mathrm dx\\ &=\int_0^\frac\pi2\left(\dfrac\pi2-x\right)\mathrm dx\\ &=\dfrac{\pi^2}{8} \end{align}
Since the problem is not entirely clear to me, please comment if it doesn't answer your question.