I'm trying to find the area of an ellipse given by $x^2 - xy + y^2 = 2$.
I am using the mapping $(u, v) = (x-\frac{y}{2}, \frac{\sqrt{3}y}{2})$, since the ellipse can be written as $(x-\frac y 2) ^2 + \frac {3y^2}{4} = 2$, so $u^2 + v^2 = 2$.
I get that $\frac {\partial (x,y)}{\partial(u,v)} = 2/\sqrt3$
And so my integral would be $\frac {2} {\sqrt{3}} \iint_D u^2 + v^2 du \, dv$.
In polar co-ordinates, this comes to $\frac {2} {\sqrt{3}} \int_0^2 \int_0^{2\pi} r^3 dr \, d\theta = \frac {16\pi}{\sqrt{3}}$.
However, the correct answer is $\frac { 4 \pi } {\sqrt3}$ given here http://tutorial.math.lamar.edu/Classes/CalcIII/ChangeOfVariables.aspx#MultInt_Change_Ex4
What is wrong with my method? I'm assuming its something to do with my original mapping because that is where my solution diverges from the one given, but I don't see why my mapping is invalid.
I checked your method and calculations: it's set up correctly, it almost works, and it is fine to use instead of the substitution in the solution given. But the error happens when you convert to polar coordinates, where you overcompensated for the different radius.
In your case, $u^2+v^2 = 2$, so you have a circle of radius $\sqrt 2$. So the endpoints of your integral with respect to $r$ should not be 0 and 2, they should be 0 and $\sqrt 2 $.