Double integral over a typical region. 
Question: Let $R$ be the region in $xy$-plane bounded by the curves $y=x^2,y=4x^2,xy=1,xy=5$.
Then find the value of line integral $\iint_R\frac{y^2}{x}dydx$.

I found it tough and time taking if I evaluate this integral from general method, is there any theorem which make it easy? Since the question is asked in competitive exam so I hope there must be an easy and fast way to solve it.
Any help will be appreciable.
Thanks in advance.
 A: Try a change of variables by letting $u=x^2/y$ and $v=xy$. Then the region $R$  is transformed into the rectangle $[1/4,1]\times [1,5]$. Since the Jacobian determinant is
$$\det\left(\begin{bmatrix}
\frac{\partial u}{\partial x}& \frac{\partial u}{\partial y}\\
\frac{\partial v}{\partial x}& \frac{\partial v}{\partial y}
\end{bmatrix}\right)
=\det\left(\begin{bmatrix}
2x/y& -x^2/y^2\\
y& x
\end{bmatrix}\right)=\frac{3x^2}{y},$$
it follows that
$$\iint_R\frac{y^2}{x}dydx=\frac{1}{3}\int_{u=1/4}^1\int_{v=1}^5\frac{y^3}{x^3}dvdu=
\frac{1}{3}\int_{1/4}^1\frac{du}{u^2}\cdot\int_{1}^5vdv.$$
Can you take it from here?
A: I would suggest change of variable as -
$\displaystyle xy = u, \frac{y}{x^2} = v$
That leads to $\displaystyle J = \frac{1}{3v}$ and integrand $\displaystyle \frac{y^2}{x} = uv$
So the integral simplifies to $\displaystyle \int_1^4 \int_1^5 \frac{u}{3} du \, dv = 12$
The intuition for change of variable comes from seeing the curves they are bound by. As you have two curves with $xy = c$ and two curves with $y = cx^2$. So you can use change of variable to integrate over a rectangular region. One of them is $u = xy$ and the other can be $\frac{x^2}{y}$ or $\frac{y}{x^2}$. I chose the latter as it simplified the integrand as $\frac{y^2}{x} = uv$.
