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Surface integrals and parametrization

  1. Mar 12, 2013 #1
    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. jcsd
  3. Mar 12, 2013 #2
    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.
     
  4. Mar 12, 2013 #3
    Oh I see

    Is r(u,v) = (v*(u^2),u) a possibility?
     
  5. Mar 12, 2013 #4
    What are the boundaries for u and v?
     
  6. Mar 12, 2013 #5
    -sqrt(6) =< u =< sqrt(6)

    I'm not sure about the v : 0 =<v=<6 ?
     
  7. Mar 12, 2013 #6

    HallsofIvy

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    [itex]x= 6- u^2[/itex], [itex]y= u[/itex] is a parameterization of the bounding curve, not the region. Myself, I would just use (x, y, 0) with [itex]0\le x\le \sqrt{6}[/itex], [itex]0\le y\le 6- x^2[/itex], using x and y themselves as parameters.
     
  8. Mar 12, 2013 #7
    I hadn't thought about that one.

    But I'm curious what would the boundaries for v be in the other parameterization?
     
  9. Mar 12, 2013 #8
    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 ##.
     
  10. Mar 12, 2013 #9
    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 :(
     
  11. Mar 12, 2013 #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.
     
  12. Mar 12, 2013 #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)

    ?
     
  13. Mar 12, 2013 #12
    Seems OK.
     
  14. Mar 12, 2013 #13
    Thank you voko.
     
  15. Mar 12, 2013 #14

    LCKurtz

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    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.
     
  16. Mar 12, 2013 #15

    But what do you mean with best idea ?

    I can see that both parameterizations can work but how is one better than the other ?
     
  17. Mar 12, 2013 #16

    LCKurtz

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    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.
     
  18. Mar 12, 2013 #17
    Okay. Thank you :smile:
     
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