# Change the order of integration $\int_{1}^{\frac{4}{3}}{\rm d}x\int_{\sqrt{\frac{x}{3}}}^{2- x}f\left ( x, y \right ){\rm d}y$

Change the order of integration $$\int_{0}^{1}{\rm d}x\int_{\sqrt{\frac{x}{3}}}^{\sqrt{x}}f\left ( x, y \right ){\rm d}y+ \int_{1}^{\frac{4}{3}}{\rm d}x\int_{\sqrt{\frac{x}{3}}}^{2- x}f\left ( x, y \right ){\rm d}y$$

By Desmos_ https://www.desmos.com/calculator/npgmmkcbnn?lang=vi, we have $$\int_{0}^{1}{\rm d}x\int_{\sqrt{\frac{x}{3}}}^{\sqrt{x}}f\left ( x, y \right ){\rm d}y= \int_{0}^{1}{\rm d}y\int_{y^{2}}^{3y^{2}}{\rm d}x$$ for the picture_ https://www.desmos.com/calculator/oy0qncqoid?lang=vi I couldn't find the inner $$\int_{1}^{\frac{4}{3}}{\rm d}x\int_{\sqrt{\frac{x}{3}}}^{2- x}f\left ( x, y \right ){\rm d}y= \int_{\sqrt{\frac{1}{3}}}^{1}{\rm d}y\int_{a}^{b}f\left ( x, y \right ){\rm d}x$$ please help me get $$a$$ and $$b$$ that $$x$$ corresponds to, and how can we solve this without drawing, which as same as what I said here_ Find the relationship between $x$ and $y$ so that $y:=0\rightarrow\frac{\pi}{2}\Leftrightarrow x:=y\rightarrow\frac{\pi}{2}$ _thanks a real lot !

Find the $$y$$ coordinates of intersection of curves $$x=y^2$$ and $$x = 3y^2$$ with line $$x+y=2$$.

$$x=2-y = y^2 \implies y = 1$$

$$x=2-y = 3y^2 \implies y = \frac{2}{3}$$

For $$0 \leq y \leq \frac{2}{3}, y^2 \leq x \leq 3y^2$$

For $$\frac{2}{3} \leq y \leq 1, y^2 \leq x \leq 2-y$$

So the integral becomes,

$$\displaystyle \int_0^{2/3} \int_{y^2}^{3y^2} f(x,y) dx \ dy + \int_{2/3}^1 \int_{y^2}^{2-y} f(x,y) dx \ dy$$