Sketch the domain D bounded by y = x^2, y = (1/2)x^2, and y=6x. Use a change of variables with the map x = uv, y = u^2 (for u ? 0) to calculate Double integral Dy^ - 1dx dy. This is an improper integral since f(x, y) = y^ - 1 is undefined at (0, 0), but it becomes proper after changing variables.

Under the given transformation, the Jacobian and its determinant are[tex]\begin{cases}x=uv\\y=u^2\end{cases}\implies J=\begin{bmatrix}v&u\\2u&0\end{bmatrix}\implies|\det J|=2u^2[/tex]so that[tex]\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\iint_{D'}\frac{2u^2}{u^2}\,\mathrm du\,\mathrm dv=2\iint_{D'}\mathrm du\,\mathrm dv[/tex]where [tex]D'[/tex] is the region [tex]D[/tex] transformed into the [tex]u[/tex]-[tex]v[/tex] plane. The remaining integral is the twice the area of [tex]D'[/tex].Now, the integral over [tex]D[/tex] is[tex]\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\left\{\int_0^6\int_{x^2/2}^{x^2}+\int_6^{12}\int_{x^2/2}^{6x}\right\}\frac{\mathrm dx\,\mathrm dy}y[/tex]but through the given transformation, the boundary of [tex]D'[/tex] is the set of equations, [tex]\begin{array}{l}y=x^2\implies u^2=u^2v^2\implies v^2=1\implies v=\pm1\\y=\frac{x^2}2\implies u^2=\frac{u^2v^2}2\implies v^2=2\implies v=\pm\sqrt2\\y=6x\implies u^2=6uv\implies u=6v\end{array}[/tex]We require that [tex]u>0[/tex], and the last equation tells us that we would also need [tex]v>0[/tex]. This means [tex]1\le v\le\sqrt2[/tex] and [tex]0<u<6v[/tex], so that the integral over [tex]D'[/tex] is[tex]\displaystyle2\iint_{D'}\mathrm du\,\mathrm dv=2\int_1^{\sqrt2}\int_0^{6v}\mathrm du\,\mathrm dv=\boxed6[/tex]