Paul's Online Notes
Home / Calculus III / Multiple Integrals / Change of Variables
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

### Section 4-8 : 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 $$R$$.

Show Solution

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!