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 !


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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,

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


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