# Find the relationship between $x$ and $y$ so that $y:=0\rightarrow\frac{\pi}{2}\Leftrightarrow x:=y\rightarrow\frac{\pi}{2}$

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

• What are the limits for your integral? $y \leq x \leq \frac{\pi}{2}$ and $0 \leq y \leq \frac{\pi}{2}$? Now you want to change order $dx \ dy$ to $dy \ dx$? Is that the question? – Math Lover Jan 17 at 10:19
• ok I see it now – Math Lover Jan 17 at 12:24
• I think to get the answer you must have to provide more details. Such as what is the main problem ? You told about a double integral, what is it ? Because it seems incomplete question from my side.@hd_30102 – nmasanta Jan 23 at 5:10
• Your previous edit does not provide any extra information ? We need detailed information about your question. – nmasanta Jan 23 at 8:20
• @haidangel I think you need to give more details. For the example by Martund's answer, do you mean you can evaluate the double integral by a new approach? In your new approach, you need to find two functions $f_1, f_2$ such that $y= f_{1}\left ( x \right ), x= f_{2}\left ( y \right )$? – River Li Feb 21 at 0:20

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: $$$$\iint\limits_{0 That, which I've written below the integral sign is the region of integration, and this integration will find out area of the region given by $$$$A = \left\{(x,y)\Bigg|0 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}
• What I want to try is set $y= f_{1}\left ( x \right ), x= f_{2}\left ( y \right ),$ but seems like no useful – haidangel Jan 24 at 6:16