# Domain of parametric plane

1. May 13, 2013

### Nikitin

1. The problem statement, all variables and given/known data
A triangle is defined by the 3 points P=(1,0,0), Q=(0,2,0) and R=(0,0,2).

Set up the double integral over its area.

3. The attempt at a solution

The triangle can be described as a plane 2x+y+z+2=0, with xE [0,1], yE [0,2], zE [0,2].

I parametrized it into r(u,v) =u*PQ+v*PR+P = u[-1,2,0]+v[-1,0,2]+(1,0,0).

But what's the domain of r(u,v)? I think it is uE [0,1-v], vE[0,1], but even if it is I found it using only gut feeling.

2. May 13, 2013

### LCKurtz

What you need to do is draw a $uv$ coordinate system. Then mark which $uv$ points are mapped to which points on your plane. For example $(u,v)=(0,0)$ goes to $(1,0,0)$. Mark also where $(u,v) = (1,0)$ and $(u,v)=(0,1)$ go. This will give you a picture of your $uv$ domain and you will be able to write your double integral limits for $u$ and $v$.

3. May 13, 2013

### Nikitin

Could you explain the general strategy in further detail? Why did you pick (u,v) = (0,0), (1,0) and (0,1), for instance? Is it because those are the minimal and maximal values of u and v?

I tried to draw the u,v plane, but I wasn't sure what to do...

4. May 13, 2013

### LCKurtz

Just draw two perpendicular axes and label them u and v instead of x and y.

If you look at your parameterization, r(u,v) =u*PQ+v*PR+P = u[-1,2,0]+v[-1,0,2]+(1,0,0) it is obvious that that those values of u and v give the corners of your triangle. Joining those the points (u,v) = (0,0), (1,0) and (0,1) in the uv plane gives you a picture of the uv domain.

It is just like as if you had written the xyz equation of the plane as z = 2-2x-y and used x and y as your parameters. Then you would have used the domain triangle in the xy plane for your xy limits.

5. May 13, 2013

### Nikitin

hmm, OK. Thanks