1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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

    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


    User Avatar
    Science Advisor

    [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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted