Finding Area Bounded by Ellipse Using (u,v) Transformation

• Master J
In summary, the student is trying to find the area of an ellipse under the condition that x=u/SQRT 2 and y=v/SQRT 3. They have found the area is 1/sqrt{6} times the area of the circle. However, they are not sure how to find the limits of integration or why the area is 1/sqrt{6} times the area of the circle.
Master J
Using a transformation (u,v) -> (x, y), I need to find the area bounded by a particular ellipse.

I am given expressions for x as u,and y as v, they are simple functions.

So I can simple plug in the expressions for u and v for the ellipse, and compute the Jacobian. But I was not given any limits of integration. I am unsure how to work them out.

Any help?

Thank you.

Let's see what you've done so far, please.

Jacobian is 1/SQRT 6

Ellipse:2x^2 + 3y^2 = 6

Ellipse, under x=u/SQRT 2 and y=v/SQRT 3

is: u^2 + v^2 = 6

It looks to me like you're doing a transformation (x,y) -> (u,v), not the other way around. Anyway, do you really even need to do an integral? The graph of the relation $u^2+v^2=6$ is just a circle in the uv plane!

Yea my bad, they were meant to be the other way round.

I know that's a circle, but I'm pretty sure we are meant to use the double integral method.

Now that I have the Jacobian and the function in terms of u and v, I am unsure how to proceed? The limits are what is stopping me!

To find the area, you want to integrate over the entire ellipse so, since the ellipse is mapped to the circle, you want to integrate over the entire circle: $-\sqrt{6}\le x\le \sqrt{6}$, $-\sqrt{6- x^2}\le y\le \sqrt{6- x^2}$. But I agree with Tom Matton. I would consider it far better to show that you understand what you are doing rather than just apply a formula: the area of the ellipse is $1/sqrt{6}$ times the area of the circle.

Actually if one of my students were to answer this by noting that the area inside the ellipse $x^2/a^2+ y^2/b^2= 1$ is $\pi ab$, I would give that full credit!

1. How do you find the area bounded by an ellipse using (u,v) transformation?

To find the area bounded by an ellipse using (u,v) transformation, you can use the formula A = πab, where a and b are the semi-major and semi-minor axes of the ellipse, respectively. These values can be determined by solving the equation u^2/a^2 + v^2/b^2 = 1, which represents the equation of an ellipse in (u,v) coordinates.

2. What is the significance of using (u,v) transformation to find the area bounded by an ellipse?

(u,v) transformation is a helpful tool in mathematics and physics, as it allows for a simpler and more efficient way of solving problems involving ellipses. By transforming the equation of an ellipse into (u,v) coordinates, we can easily find the area bounded by the ellipse using the aforementioned formula, without having to consider the complex algebraic equations involved in traditional methods.

3. Can (u,v) transformation be used for any type of ellipse?

Yes, (u,v) transformation can be used for any type of ellipse, as long as its equation can be expressed in terms of u and v. This includes both standard and non-standard ellipses, as well as tilted and rotated ellipses.

4. Are there any limitations to using (u,v) transformation to find the area bounded by an ellipse?

One limitation is that (u,v) transformation only applies to ellipses in two dimensions. It cannot be used to find the area of an ellipse in three-dimensional space. Additionally, if the equation of the ellipse is not given in terms of u and v, then (u,v) transformation cannot be applied.

5. How is (u,v) transformation related to polar coordinates?

(u,v) transformation is closely related to polar coordinates, as both involve transforming an equation from Cartesian coordinates to a different coordinate system. In fact, polar coordinates can be seen as a special case of (u,v) transformation, where a and b are equal. This makes (u,v) transformation a more general and versatile method for solving problems involving ellipses.

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