TITLE: Integrating over an area
QUESTION [5 upvotes]: I need to integrate $f(x,y):=x^2y^2$ over an area $B\subset\mathbb R^2$that is restricted by the following 4 functions: $$y=\frac x9;\;y=\frac x4;\;y=\frac 1x;\;y=\frac4x;$$
Of course due to the symmetry we can integrate just over the area in the positive $x$-$y$ region and then multiply the result by two. 
I am confused by choosing the boundaries for integration. On the one hand it could be $$\frac x9 \leq y\leq \frac x4,\;2\leq x\leq6$$ 
or alternatively $$\frac 1x \leq y\leq \frac 4x,\;2\leq x\leq6$$
So how do I choose the boundaries to calculate $\int_Bx^2y^2d\mu(x,y)?$

REPLY [3 votes]: Switch to polar coordinates. Then you have the following integral:
$$\int_{\tan^{-1}\frac{1}{9}}^{\tan^{-1}\frac{1}{4}} \int_{\sqrt{2\csc(2\theta)}}^{2\sqrt{2\csc(2\theta)}}r^5\sin^2\theta\cos^2\theta\,dr\,d\theta=\frac{1}{6}\int_{\tan^{-1}\frac{1}{9}}^{\tan^{-1}\frac{1}{4}}\sin^2\theta\cos^2\theta (504\csc^3(2\theta))\,d\theta $$
$$=\frac{504}{6}\int_{\tan^{-1}\frac{1}{9}}^{\tan^{-1}\frac{1}{4}}\frac{\sin^2\theta\cos^2\theta}{8\sin^3\theta \cos^3\theta}\,d\theta=\frac{504}{24}\int_{\tan^{-1}\frac{1}{9}}^{\tan^{-1}\frac{1}{4}}\csc(2\theta)\,d\theta$$
$$=\boxed{21\ln\left(\frac{3}{2}\right)}$$