Evaluate the integral e^xy w region d xy=1, xy=4, x/y=1, x/y=2

Make a change of coordinates:

[tex]u(x,y)=xy[/tex]
[tex]v(x,y)=\dfrac xy[/tex]

The Jacobian for this transformation is

[tex]\mathbf J=\begin{bmatrix}\dfrac{\partial u}{\partial x}&\dfrac{\partial v}{\partial x}\\\\\dfrac{\partial u}{\partial y}&\dfrac{\partial v}{\partial y}\end{bmatrix}=\begin{bmatrix}y&x\\\\\dfrac1y&-\dfrac x{y^2}\end{bmatrix}[/tex]

and has a determinant of

[tex]\det\mathbf J=-\dfrac{2x}y[/tex]

Note that we need to use the Jacobian in the other direction; that is, we've computed

[tex]\mathbf J=\dfrac{\partial(u,v)}{\partial(x,y)}[/tex]

but we need the Jacobian determinant for the reverse transformation (from [tex](x,y)[/tex] to [tex](u,v)[/tex]. To do this, notice that

[tex]\dfrac{\partial(x,y)}{\partial(u,v)}=\dfrac1{\dfrac{\partial(u,v)}{\partial(x,y)}}=\dfrac1{\mathbf J}[/tex]

we need to take the reciprocal of the Jacobian above.

The integral then changes to

[tex]\displaystyle\iint_{\mathcal W_{(x,y)}}e^{xy}\,\mathrm dx\,\mathrm dy=\iint_{\mathcal W_{(u,v)}}\dfrac{e^u}{|\det\mathbf J|}\,\mathrm du\,\mathrm dv[/tex]
[tex]=\displaystyle\frac12\int_{v=}^{v=}\int_{u=}^{u=}\frac{e^u}v\,\mathrm du\,\mathrm dv=\frac{(e^4-e)\ln2}2[/tex]