Area of Tilted Ellipse?

Following your simplification after a change of variables X = x + 2y + 1 , Y = y - 2, we can simply integrate dXdY over the XY region and multiply it by the Jacobian.

The Jacobian (as I have been taught) is the determinant of the following matrix (abbreviated J):

[ ∂ x/ ∂ X .....∂x/∂ Y ]

[ ∂ y/ ∂ X......∂y/∂ Y]

and is the measure of how much the new ellipse is "streched" at any given point. The total integral gets multiplied by this in the new coordinate system. In other words dxdy = JdXdY.

We solve for x and y to get,

x = X - 2Y - 5

y = Y + 2

and J =

| 1 ... -2|

| 0.... 1.| = 1

How surprising! This means dxdy = dXdY so that the integral is the same! But we can't simply generalize this to ANY tilted ellipse. What's interesting though is that any region in the xy plane has the same area after being transformed to XY plane!

I hope this makes sense.

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Northstar, yes, it does mean that Jacobian has to be 1 for this to work.

Scythian has done a nice generalization of this! But we have to be careful what we call Jacobian. For the tranformation (x,y) -> (X,Y) the Jacobian is 1/(ae - bd) while for (X,Y) mapping to (x,y) Jacobian is ae - bd.

The semi-axes of your ellipse are actually 1.14001 and 5.26311, but nevertheless the area comes out to (1.14001)(5.26311)π = 6π. It's an affine transform. If the ellipse intersects the line y = 2 at two points x = A + 3 and x = A - 3, for some value A, and if there's a double tangent at two y = B + 2 and y = B - 2, for some value B, then the area of the ellipse is the same as an ellipse of semimajor axes of 3 and 2. This would only work if both terms of the expression X²/A² + Y²/B² = 1 are linear in both x and y, such as (x + 2y + 1) and (y - 2).

Edit: Any ellipse expressed by the equation:

(ax + by + c)² / A² + (dx + ey + f)² / B² = 1

has the area

ABπ / (ae - bd)

Hence, for your equation, a = 1, b = 2, d = 0, e = 1, the area is ABπ. This agrees with Zeta's answer below. This result was worked out in full by brute integration.

Edit 2: Yes, Zeta is saying that the Jacobian must equal 1 for this to work. The expression (ae - bd) IS the Jacobian in the case of 2D ellipses.

Edit 3: Uh, the absolute value of (ae - bd) should be taken. After all, if the limits of the integration is done the other way, you can get a negative area, but nobody looks at areas that way.

Your answer is correct, however I am slightly surprised, because the semi-major axis is not 3, but rather a=5.263114932 and the semi-minor axis is not 2, but rather b=1.140009306. Nonetheless, the area is

area = πab = π*5.263114932*1.140009306 = 6π

I know this because I worked out a, b the hard way. Why your form works is puzzling and I will have to give it more though to give you a proper explanation.

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EDIT: I connected here this morning to add a note about Jacobian, similar to that of zeta and Scythian -- but I see it is already here so there is no need. This is basic linear algebra or calculus and I should have realized it before. The Jacobian determinant measures area (or volume) change (and orientation) and for an affine or linear transformation it is constant. It is interesting that you can find the area of slanted ellipses without actually rotating them or determining the lengths of their axes.

As you've been told by other answers, no. Heisenberg Uncertainty Principle comes from being able to calculate the position OR the momentum of a sub-atomic particle, but not both at the same time. The two answers you get from a Quadratic are both definitive answers, and really don't have much to do with the movement of subatomic particles.