Changing integration order in an iterated double integral

Question

I am trying to solve the integral below (equation 1) by changing the integration order from dydx to dxdy:

$$ I = \int_{\frac{2}{\sqrt{5}}}^1\int_{\sqrt{1-x^2}}^{\frac{x}{2}}f(x,y) \mathrm dy \mathrm dx + \int_1^{\frac{4}{\sqrt{5}}}\int_0^{\frac{x}{2}} f(x,y) ​\mathrm dy \mathrm dx+\int_{\frac{4}{\sqrt{5}}}^2\int_0^{\sqrt{4-x^2}} f(x,y) ​\mathrm dy \mathrm dx\tag{1} $$

This is the plot of the integration region:

The function f(x) may be considered arbitrary. I was trying to rewrite the integration boundaries but I have not managed to do it correctly. I would be most grateful if you may help me.

Answer 1

Please see the shaded region for integration in the diagram.

As you can see, bounds are easier in polar coordinates.

$1 \leq r \leq 2$

$x \geq 2 y \implies 0 \leq \theta \leq \arctan \left(\frac{1}{2}\right)$

$\displaystyle \int_0^{\arctan \left(\frac{1}{2}\right)} \int_1^2 r \ f(x = r \cos\theta, y = r \sin\theta) \ dr \ d\theta$

But if you have to change the order of integral from $dy \ dx$ to $dx \ dy$,

Line $x = 2y$ intersects with $x^2+y^2 = 1$ at $A \left (\cfrac{2}{\sqrt5}, \cfrac{1}{\sqrt5} \right)$ and with $x^2 + y^2 = 4$ at $B \left (\cfrac{4}{\sqrt5}, \cfrac{2}{\sqrt5} \right)$.

In the order $dy \ dx$,

i) for $0 \leq y \leq \cfrac{1}{\sqrt5}$, $x$ is bound between circles $r = 1$ and $r = 2$ i.e. $\sqrt{1-y^2} \leq x \leq \sqrt{4-y^2}$.

ii) for $\cfrac{1}{\sqrt5} \leq y \leq \cfrac{2}{\sqrt5}$, $x$ is bound between line $x = 2y$ and circle $r = 2$ i.e. $2y \leq x \leq \sqrt{4-y^2}$

Answer 2

In the region of the plane bounded by the ray $[0,\infty)$ and the ray $y=\frac x2$ ($x\leqslant 0$), we are dealing with any pair $(x,y)$ such that $1\leqslant x^2+y^2\leqslant4$. If $x\leqslant\frac1{\sqrt5}$, $x$ can take any value such that $(x,y)$ lies between the first circular arc from the picture below to the second one (that is, $\sqrt{1-y^2}\leqslant x\leqslant\sqrt{4-y^2}$) and otherwise $x$ can take any value such that $(x,y)$ lies between the line $x=2y\left(\iff y=\frac x2\right)$ and the second circular arc. So, consider$$\int_0^{1/\sqrt5}\int_{\sqrt{1-y^2}}^{\sqrt{4-y^2}}f(x,y)\,\mathrm dx\,\mathrm dy+\int_{1/\sqrt5}^1\int_{2y}^{\sqrt{4-y^2}}f(x,y)\,\mathrm dx\,\mathrm dy.$$