# 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

Staff Emeritus
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!