# Surface integrals and parametrization

1. Mar 12, 2013

### Tala.S

An area A in the xy-plane is defined by the y axis and by the parabola with the equation
x=6-y^2.

Furthermore a surface S is given by that part of the graph for the function h(x,y)=6-x-y^2 that satisfies x>=0 and z>=0.

I have to parametrisize A and S.

Could this be a parametrization of A :

(6-u^2,u,0) where u = y

?

I find this subject extremely hard to understand so I will appreciate any kind of help.

2. Mar 12, 2013

### voko

A is two-dimensional. Consequently, it takes two variables to parametrize it. So your parametrization cannot be correct.

Usually, one wants to parametrize in such a way that the parameters' domain is "nice", e.g., a rectangle.

3. Mar 12, 2013

### Tala.S

Oh I see

Is r(u,v) = (v*(u^2),u) a possibility?

4. Mar 12, 2013

### voko

What are the boundaries for u and v?

5. Mar 12, 2013

### Tala.S

-sqrt(6) =< u =< sqrt(6)

I'm not sure about the v : 0 =<v=<6 ?

6. Mar 12, 2013

### HallsofIvy

Staff Emeritus
$x= 6- u^2$, $y= u$ is a parameterization of the bounding curve, not the region. Myself, I would just use (x, y, 0) with $0\le x\le \sqrt{6}$, $0\le y\le 6- x^2$, using x and y themselves as parameters.

7. Mar 12, 2013

### Tala.S

But I'm curious what would the boundaries for v be in the other parameterization?

8. Mar 12, 2013

### voko

Assuming $y = u$, $-\sqrt{6} \le u \le \sqrt {6}$ is OK.

But you still have to have a range for v. It can be arbitrarily taken to be $0 \le v \le 1$.

Now you have to have $x = f(u, v)$. For this you could consider the boundaries: $x = 0$ and $x = 6 - y^2$. You could further simplify this by checking out $f(u, v) = g(u)v$.

9. Mar 12, 2013

### Tala.S

I don't understand why you choose the boundaries for v to be that.

And the last two lines.....wasn't x = v(6-u^2) with the boundaries -√6 ≤ u ≤ √6.

I'm a bit confused now :(

10. Mar 12, 2013

### voko

Parametrization is always arbitrary. If you have some parametrization X = P(Y), you can also have Y = Q(Z), which gives you parametrization X = R(Z) = P(Q(Z)). Functions P and Q, as well as their domains, are arbitrary. There are infinitely many ways to parametrize.

As I remarked above, one usually wants a parametrization that is "nice" or at least simple in a certain way.

I chose the domain to be rectangular. Why rectangular? No particular reason in this case, but when you need, for example, to integrate something, it might simplify things.

$x = v(6 - u^2)$ is good. No, you did not have it this way earlier.

11. Mar 12, 2013

### Tala.S

The explanation was good. Thank you.

I have one question concerning the parametrization of S.

since z=h(x,y) wouldn't a possible parametrization of S be :

r(u,v) = (v(6-u^2), u, 6-(6-u^2)*v-u^2)

?

12. Mar 12, 2013

### voko

Seems OK.

13. Mar 12, 2013

### Tala.S

Thank you voko.

14. Mar 12, 2013

### LCKurtz

OK, I can't resist chiming in on this one. For the region A, Halls' idea is best except he has the variables reversed. For A the natural parameterization is $x=x,\ y=y,\ z = 0$ giving $r(x,y) = \langle x,y,0\rangle,\ 0\le x\le 6-y^2,\ -\sqrt 6\le y\le \sqrt 6$.

For the surface, the natural parameterization would be almost the same, except for $z$:$$r(x,y) = \langle x,y, 6-x-y^2\rangle$$with the same limits.

15. Mar 12, 2013

### Tala.S

But what do you mean with best idea ?

I can see that both parameterizations can work but how is one better than the other ?

16. Mar 12, 2013

### LCKurtz

It is "best" in my opinion because it is simplest and there is no reason to use anything more complicated in this problem. If the domain were, for example, a circle, I would give a different parameterization because of that.

17. Mar 12, 2013

### Tala.S

Okay. Thank you