Calculus III - Change of Variables

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Section 15.8 : Change of Variables

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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!