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Section 15.8 : Change of Variables

4. If \(R\) is the region inside \(\displaystyle \frac{{{x^2}}}{4} + \frac{{{y^2}}}{{36}} = 1\) determine the region we would get applying the transformation \(x = 2u\), \(y = 6v\) to \(R\).

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There really isn’t a lot to this problem.

It should be pretty clear that the outer boundary of \(R\) is an ellipse. That isn’t really important to this problem but this problem will lead to seeing how to set up a nice transformation for elliptical regions.

To determine the transformation of this region all we need to do is plug the transformation boundary equation for \(R\). Doing this gives,

\[\frac{{{{\left( {2u} \right)}^2}}}{4} + \frac{{{{\left( {6v} \right)}^2}}}{{36}} = 1\hspace{0.25in}\to \hspace{0.25in}\frac{{4{u^2}}}{4} + \frac{{36{v^2}}}{{36}} = 1\,\hspace{0.25in}\to \hspace{0.25in}{u^2} + {v^2} = 1\]

So, the boundary equation for \(R\) transforms into the equation for the unit circle and so, under this transformation, we can transform an ellipse into a circle (a unit circle in fact…).

You can see how to determine a transformation that will transform an elliptical region into a circular region can’t you? Integrating over an elliptical region would probably be pretty unpleasant but integrating over a unit disk will probably be much nicer so this is a nice transformation to understand how to get!