Results 1 to 2 of 2

Math Help - Surface area

  1. #1
    Member
    Joined
    May 2010
    Posts
    241

    Post Surface area

    Hi there, sorry for the hurry, I had a big problem this weekend and I couldn't study. I have my exam tomorrow, and need help with this. I'm finishing with the topic, and I have almost studied all what I had to study, but this exercise resulted problematic.

    This is it.

    It asks me to calculate the area of the surface x^2-y^2-z^2=0 under the conditions x\geq{0},y\geq{0},z\geq{0},x\leq{1-z}

    So I've made the parametrization this way:
    \begin{Bmatrix}x=u\\y=u \cos v \\z=u \sin v\end{matrix}
    And I've found: ||T_u\times{T_v}||=\sqrt[ ]{2}u

    Then I made the parametrization for the plane:
    \begin{Bmatrix}x=u\\y=v \\z=1-u\end{matrix}

    Then I made the integral for the surface, well, I've tried:
    \displaystyle\int_{0}^{2\pi}\displaystyle\int_{0}^  {1-u}\sqrt[ ]{2}u dudv

    I think that the surface is actually not bounded.

    Actually, when I did the intersection I've found a parabola:
    \sqrt[ ]{y^2+z^2}=1-z\longrightarrow{y^2+2z-1=0}

    Which I think makes sense, but the surface would be infinite, and I don't know what to do.

    Any help will be thanked.
    Last edited by mr fantastic; November 22nd 2010 at 05:26 PM. Reason: Urgent is irrlevant to the post title.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Nov 2010
    Posts
    6
    Quote Originally Posted by Ulysses View Post
    Then I made the integral for the surface, well, I've tried:
    \displaystyle\int_{0}^{2\pi}\displaystyle\int_{0}^  {1-u}\sqrt[ ]{2}u dudv

    I think that the surface is actually not bounded.

    Actually, when I did the intersection I've found a parabola:
    \sqrt[ ]{y^2+z^2}=1-z\longrightarrow{y^2+2z-1=0}
    If you look at your integral, the limits are actually lines on the plane, right? Try using those as bounds and compare to where the parabola comes across them.
    So you would have your parabola on the uv-plane bounded by:
    v = 0; v = 1 - u; u = 0; u = 2pi. Something to note: since you have your integral limit set up as a u to v, your integral should reflect this with: sqrt(2)udvdu. Order is important. This intersected area should be the region you're finding the surface area for.

    Hope this helps, and if anywhere I am mistaken, someone please feel free to correct me and get him pointed in the right direction! =)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Surface Area of Surface
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 1st 2010, 09:53 AM
  2. Calculate the surface area of the surface
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 26th 2009, 04:03 AM
  3. Lateral Area and Total Surface Area
    Posted in the Geometry Forum
    Replies: 1
    Last Post: May 25th 2009, 04:28 PM
  4. Help finding surface area of a surface
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 3rd 2008, 04:11 PM
  5. Volume, Surface Area, and Lateral Surface Area
    Posted in the Geometry Forum
    Replies: 1
    Last Post: April 14th 2008, 11:40 PM

Search Tags


/mathhelpforum @mathhelpforum