# Integral over a Rotated Ellipse

• SpringPhysics
In summary, the student is trying to solve for the integral of f(x,y)=xy over x^2 - xy + 2y^2 = 1 in terms of an integral over the unit circle, but does not know how to start the problem or find the angle of rotation.

## Homework Statement

Determine the integral of f(x,y)=xy over x^2 - xy + 2y^2 = 1 in terms of an integral over the unit circle.

## The Attempt at a Solution

The associated hint is to complete the square (which I did...and got messy expressions for x and y). I would like to know how to start the problem, since changing directly to polar coordinates will not work. I have tried to use the rotation change of variables (after which I can use polar coordinates), but I do not know the angle of rotation nor the lengths of the semi- axes.

Is there another change of variables that I am overlooking?

Thanks.

This page tells you how to find the angle of rotation to get rid of the cross term.

http://www.sparknotes.com/math/precalc/conicsections/section5.rhtml

Or you could derive it yourself by using the rotation formulas

\begin{align*} x &= x' \cos \theta - y' \sin \theta \\ y &= x' \sin \theta + y' \cos\theta \end{align*}

Substitute in for x and y and solve for the angle that causes the cross term to vanish.

I solved the tedious resulting system of equations and got 3=2...which means that there is no such rotation?

EDIT: I get the angle of rotation as 0 when solving for the coefficient of xy to be 0, but according to the formula from the link, the angle should be pi/8.

EDIT2: I get -pi/8 for the angle of rotation, as opposed to the pi/8 by using (A-C)/B.

I did some research and found the change of coordinates related to the shear factor. I believe this makes more sense because the x-intercepts of the ellipse corresponds to the x-intercepts of the unit circle. So now it's a question of how to determine where the vector (0,1) is mapped to on the ellipse.

On second thought, I'll just stick with the rotation and scaling.