I would probably try to tackle it by changing variables to get it into a circle. I think I know how I would first try to do it, but I don't know if you'd know the method (and I don't remember if there's an easier, more straightforward way to see the transformation). Basically, I would write the left hand side of the equation for your ellipse as
\mathbf{r}^{\mathsf{T}}\mathsf{A}\mathbf{r} = 1
where \mathbf{r} is a column vector with entries x and y and \mathsf{A} is a 2 x 2 matrix. I would then find the eigenvalues of the matrix A and diagonalize it. Once diagonalized you might recall that you can write \mathsf{A} = \mathsf{P}^{-1}\mathsf{D}\mathsf{P}, where D is a diagonal matrix. Then define \mathbf{r}' = P\mathbf{r}. This defines news coordinates with the x' and y' axes along the major and minor axes of the ellipse. i.e., the equation for the ellipse in terms of x' and y' will be
\left(\frac{x'}{a}\right)^2 + \left(\frac{y'}{b}\right)^2 = 1,
where a and b are constants determined by your transformation. If you don't know the linear algebra to do all this, basically what you're trying to do is find a transformation
x' = Ax + By,
y' = Cx + Dy,
such that you get the above equation for the ellipse in terms of x' and y' (i.e., no xy, or x or y terms). Once you get to that point, you can define new variables u = (x'/a) and v = y'/b, which gives you the equation for a circle.
You then need to find the Jacobian to change variables in the integral to an integral over u and v.
Uh... was that helpful at all for you? Should I try to think of something simpler?
