# Area, Jacobians etc.

1. Feb 9, 2009

### 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.

2. Feb 9, 2009

### Tom Mattson

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

3. Feb 9, 2009

### Master J

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

4. Feb 9, 2009

### Tom Mattson

Staff Emeritus
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!

5. Feb 9, 2009

### Master J

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

I know thats 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!

6. Feb 9, 2009

### HallsofIvy

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!