Integrating over an area 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 $\int_Bx^2y^2d\mu(x,y)?$

 A: You need to use a nonstandard change of coordinates to solve this. Adopt $u = xy$ and $v = x/y$. 
The motivation for this is as follows: you can rewrite the equations $y=1/x$ and $y=4/x$ as $xy =1$ and $xy=4$, therefore we obtain the change $u=xy$.
You can rewrite the other two equations $y=x/4$ and $y=x/9$ as $x/y=4$ and $x/y=9$, leading to $v=x/y$. Isolating $x$ and $y$ in terms of $u$ and $v$ as defined you get $$x = \sqrt{uv}, \quad y = \sqrt{\frac{u}{v}}.$$ Computing the Jacobian and taking the absolute value gives $$|J(u,v)| = \frac{1}{2v}.$$ The new limits are $1 \leq u \leq 4 $ and $4 \leq v \leq 9$. The function $f(x,y) = x^2y^2$ is now $g(u,v) = u^2$ and the integral is $$I = \frac{1}{2} \int_1^4 \int_4^9 \frac{u^2}{v} \, dv \, du.$$ My computation yields $$I = 21 \ln \left( \frac{3}{2} \right).$$
A: Switch to polar coordinates. Then you have the following integral:
$$\int_{\tan^{-1}\frac{1}{9}}^{\tan^{-1}\frac{1}{4}} \int_{\sqrt{2\csc(2\theta)}}^{2\sqrt{2\csc(2\theta)}}r^5\sin^2\theta\cos^2\theta\,dr\,d\theta=\frac{1}{6}\int_{\tan^{-1}\frac{1}{9}}^{\tan^{-1}\frac{1}{4}}\sin^2\theta\cos^2\theta (504\csc^3(2\theta))\,d\theta $$
$$=\frac{504}{6}\int_{\tan^{-1}\frac{1}{9}}^{\tan^{-1}\frac{1}{4}}\frac{\sin^2\theta\cos^2\theta}{8\sin^3\theta \cos^3\theta}\,d\theta=\frac{504}{24}\int_{\tan^{-1}\frac{1}{9}}^{\tan^{-1}\frac{1}{4}}\csc(2\theta)\,d\theta$$
$$=\boxed{21\ln\left(\frac{3}{2}\right)}$$
A: EDIT: The first part is the calculation of the area B, the second part is the integration of function $f(x,y)$ in the area B.
The intersection of $y_1=\frac x4$ with $y_2=\frac4x$ is at $x=4,y=1$.
The intersection of $y_3=\frac x9$ with $y_4=\frac 1x$ is at $x=3,y=1/3$.
So the area is given by:
$$B=\int_{2}^{4}y_1(x) dx+\int_{4}^{6} y_2(x) dx-\int_{2}^{3}y_4(x) dx-\int_{3}^{6} y_3(x) dx$$
$$=\int_{2}^{4}\frac x4 dx+\int_{4}^{6} \frac4x dx-\int_{2}^{3}\frac 1x dx-\int_{3}^{6} \frac x9 dx=\ln\left(\frac{27}{8}\right)$$
EDIT:
Now I add the part of the integration of function $f(x,y)$ in the area B.
$$\int\int_B f(x,y)dxdy=\int_{2}^3x^2 \left(\int_{y_4(x)}^{y_1(x)} y^2 dy\right)dx+\int_{3}^4x^2 \left(\int_{y_3(x)}^{y_1(x)} y^2 dy\right)dx+\int_{4}^6 x^2 \left(\int_{y_3(x)}^{y_2(x)} y^2 dy\right)dx=21\ln(3/2)$$
