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

User Ellene
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1 Answer

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Make a change of coordinates:


u(x,y)=xy

v(x,y)=\frac xy

The Jacobian for this transformation is


\mathbf J=\begin{bmatrix}(\partial u)/(\partial x)&(\partial v)/(\partial x)\\\\(\partial u)/(\partial y)&(\partial v)/(\partial y)\end{bmatrix}=\begin{bmatrix}y&x\\\\\frac1y&-\frac x{y^2}\end{bmatrix}

and has a determinant of


\det\mathbf J=-\frac{2x}y

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


\mathbf J=(\partial(u,v))/(\partial(x,y))

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


(\partial(x,y))/(\partial(u,v))=\frac1{(\partial(u,v))/(\partial(x,y))}=\frac1{\mathbf J}

we need to take the reciprocal of the Jacobian above.

The integral then changes to


\displaystyle\iint_{\mathcal W_((x,y))}e^(xy)\,\mathrm dx\,\mathrm dy=\iint_{\mathcal W_((u,v))}(e^u)/(|\det\mathbf J|)\,\mathrm du\,\mathrm dv

=\displaystyle\frac12\int_(v=)^(v=)\int_(u=)^(u=)\frac{e^u}v\,\mathrm du\,\mathrm dv=\frac{(e^4-e)\ln2}2
User Timvp
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