# Change of variables

1. Nov 14, 2009

### Dustinsfl

Let R be the elliptical regin in the first quadrant by x^2/3^2+y^2/2^2=1. x=3u, y=2v, evaluate the area of R.

How do I setup the double integral? By this I mean what do I put in for f(g(u,v),h(u,v))jacobian dudv?

I know the bounds of u and v and the Jacobian but I not sure what to do for g and h. Would that be just pluging 3u and 2v into x^2 and y^2, respectively? But what happens to the 1 in the eqaution?

So by solving the (u,v) cordinates, one comes up with (0,0); (1,0); and (0,1).
The Jacobian is 6.

2. Nov 14, 2009

### LCKurtz

When you substitute your u's and v's for the x's and y's of the equation, for your region A in the xy plane you get a new region B in the uv plane. The point of the substitution is that instead of integrating

$$Area = \int \int_A 1\ dydx$$

over the region A in the xy plane, which is "ugly", you can instead calculate

$$Area = \int \int_B 1 J\ dudv$$

using u-v limits in the uv plane. If your new region B is nice, you may easily work out the new integral, or even avoid working it at all.

3. Nov 14, 2009

### Dustinsfl

So the integral is just 0<=u<=1 and 0<=v<=-u+1 dvdu with no equation in the the integral?

4. Nov 14, 2009

### LCKurtz

No. You need to draw a graph of your uv equation in the uv plane and use it to figure out correct limits.

5. Nov 14, 2009

### Dustinsfl

The drawing is a triangle and those are the bounds. I am not concerned about them. I am trying to figure out what f(g(u,v),h(u,v)) are inside the integrals. Is it 1? Is it u+v not considering the 1 or is it u+v-1?

Last edited: Nov 14, 2009
6. Nov 14, 2009

### LCKurtz

Are you trying to tell me that when you substitute x=3u, y=2v into the equation:

$$\frac {x^2}{3^2}+\frac {y^2}{2^2}=1$$

that the graph you get in the uv plane is a triangle???

Your integrand, once you figure out the proper area, will be 1 times the Jacobian.