Surface integrals and parametrization

Tala.S
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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.
 
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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.
 
Oh I see

Is r(u,v) = (v*(u^2),u) a possibility?
 
What are the boundaries for u and v?
 
-sqrt(6) =< u =< sqrt(6)

I'm not sure about the v : 0 =<v=<6 ?
 
Tala.S said:
Oh I see

Is r(u,v) = (v*(u^2),u) a possibility?
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.
 
I hadn't thought about that one.

But I'm curious what would the boundaries for v be in the other parameterization?
 
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 ##.
 
voko said:
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 ##.

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
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
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
Seems OK.
 
  • #13
Thank you voko.
 
  • #14
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
LCKurtz said:
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.
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
Tala.S said:
But what do you mean with best idea ?

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

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
Okay. Thank you :smile:
 
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