# Double integral with transformation

1. Jul 25, 2008

### glog

1. The problem statement, all variables and given/known data

$$\int\int x^2 dA$$ where the area D is boundd by the ellipse 5x^2 + 4xy + y^2 = 1.

2. Relevant equations

3. The attempt at a solution

I'm not sure where to start this question. A few ideas I'm exploring are: 1. rewrite in polar co-ordinates (not sure how to write an ellipse in this form!), or 2. do a transformation from this ellipse to a circle (not sure how to do this either).

I anyone can suggest which method is more correct and get me started with a general idea of how to approach the problem, i would greatly appreciate it!

Thank you.

2. Jul 25, 2008

### Mute

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?

3. Jul 25, 2008

### glog

Yeah I definitely think that change of variables is reasonable, I just think there's any easier way to turn this into a circle than this matrix approach. Hmmm...