Converting to polar double integral with a specific order of integration We are asked to write down the integral
$$I=\int_0^1 \int_x^{x\sqrt3}f(\sqrt{x^2+y^2})dydx$$
in a polar form first with respect to $\theta$ and then with respect to $r$.
I have recognised that $0<r<2$ and $r=\frac{1}{\cos\theta}$ but I am finding it difficult to find bounds for $\theta$ as a function of $r$.

The boundaries are the the three lines of the triangle, which transform in polar to
$$\begin{cases}y = x \\ y = \sqrt{3}x \\ x = 1\end{cases} \implies \begin{cases}\theta = \frac{\pi}{4} \\ \theta = \frac{\pi}{3} \\ r = \sec\theta\end{cases}$$
giving us the integral
$$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\int_0^{\sec\theta}f(r)\:r\:dr\:d\theta$$
To swap the integral, we can either look at the Cartesian graph of the lines

where $\theta$ integrals move along concentric circles, or in an $r\theta$ plane

where $\theta$ integrals move in horizontal lines parallel to the $\theta$ axis. In both cases, we can see that region is split in two by the red line, which represents the circle $r^2=2$. This gives us the two integrals
$$\int_0^{\sqrt{2}}\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}f(r)\:r\:d\theta\:dr + \int_{\sqrt{2}}^2\int_{\sec^{-1}(r)}^{\frac{\pi}{3}}f(r)\:r\:d\theta\:dr$$
$$ = \int_0^{\sqrt{2}}\frac{\pi r}{12}f(r)\:dr + \int_{\sqrt{2}}^2\left(\frac{\pi}{3}-\cos^{-1}\left(\frac{1}{r}\right)\right)rf(r)\:dr$$
