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Integral over a Rotated Ellipse

  • #1

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.


Homework Equations





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.
 

Answers and Replies

  • #2
vela
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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

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

Substitute in for x and y and solve for the angle that causes the cross term to vanish.
 
  • #3
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.

Thanks for your help.
 
Last edited:

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