Calculate $\iint\limits_Gf(x;y)$ where $$ f(x;y)=\frac{y}{x},\ \ G=\{x\geqslant0,\ y\geqslant0,\ 1\leqslant xy\leqslant 5,\ \frac{x}{2}\leqslant y\leqslant2x\} $$
I tried to plot the function that $G$ provides and got something like this:
However, I don't think that it is easy to find boundaries for $x$ and $y$ in order to convert given double integral into an iterated one. So, is there any better solution to this problem?
On
In polar coordinates, the region is
$$1\le r^2\sin\theta\cos\theta\le 5, \>\>\>\>\>\frac12\le \tan\theta \le 2$$
and $f(r,\theta)=\tan\theta$. So, with the limits,
$$\theta_1= \tan^{-1}\frac12,\>\>\>\>\> \theta_2= \tan^{-1}2, \>\>\>\>\>r_1^2(\theta)=\frac1{\sin\theta\cos\theta}, \>\>\>\>\>r_2^2(\theta)=\frac5{\sin\theta\cos\theta}$$
the integral is
$$I= \int_{\theta_1}^{\theta_2} \int_{r_1(\theta)}^{r_2(\theta)} \tan\theta rdrd\theta =2 \int_{\theta_1}^{\theta_2} \frac{\tan\theta}{\sin\theta\cos\theta}d\theta$$ $$=2\int_{\theta_1}^{\theta_2} \sec^2\theta d\theta =2\tan\theta|_{\theta_1}^{\theta_2}=2(2-\frac12)=3$$
Let $u(x,y) = xy, v(x,y) = y/x$ and use the change of variables theorem.