Let $D$ be the region bounded by $y = x^2, y = \frac12 x^2$ and $y = 6x$. I want to find the following $$\iint_D \frac1y dy\ dx.$$
Note that $\frac 1y$ is not defined at $(0,0)$, so I apply a change of variables to convert into a proper integral. Let $x = uv$ and $y = u^2$.
Calculating the Jacobian we have $dx\ dy = 2u^2 du\ dv$. Next, I am getting confused with the limits of the integral. Need some help!
You can compute the integral directly as $$\int_0^6\int_{\frac{1}{2}x^2}^{x^2}\frac{1}{y}dydx+\int_6^{12}\int_{\frac{1}{2}x^2}^{6x}\frac{1}{y}dydx=6\ln(2)+\int_6^{12}\left(\ln(6x)-\ln\left(\frac{1}{2}x^2\right)\right)dx.$$