1. Not finding help here? Sign up for a free 30min 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!

Different approach on calculating flux threw area

  1. Aug 13, 2009 #1
    calculate
    [tex]
    \iint_{M}^{}\vec{F}\vec{dS}
    [/tex]
    where
    [tex]
    \vec{F}=(e^y,ye^x,x^2y)
    [/tex]
    M is a part of hyperboloid [tex]x^2+y^2[/tex]
    which is located at 0<=x<=1 and 0<=y<=1 ,and its normal vector points outside
    like this :
    http://i28.tinypic.com/f9p63r.gif

    i am used to solve it like this
    [tex]
    \iint_{M}^{}\vec{F}\vec{dS}=\iint_{D}\frac{\vec{F}\cdot
    \vec{N}}{|\vec{N}\cdot\vec{K}|}dxdy
    [/tex]
    [tex]
    \vec{N}=(2x,2y,-1)
    [/tex]
    [tex]
    \iint_{M}^{}\vec{F}\vec{dS}=\iint_{D}\frac{(e^y,ye^x,x^2y) \cdot
    (2x,2y,-1)}{1}dxdy
    [/tex]
    now i convert into polar coordinates

    x^2+y^2=r
    [tex]
    =\int_{0}^{2\pi}\int_{0}^{1}\frac{(e^y,ye^x,x^2y) \cdot (2x,2y,-1)}{1}rdrd\theta
    [/tex]
    how to what are the intervals for r
    i just guessed its from 0 to 1
    i dont know how to know the upper interval here

    except that
    is this method ok?
     
  2. jcsd
  3. Aug 13, 2009 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    [itex]x^2+y^2[/itex] is just a meaningless expression and does not describe a hyperboloid. Do you mean [itex]z^2=x^2+y^2[/itex]?

    You mean [itex]x^2+y^2=r^2[/itex], right?

    Well, [itex]x^2+y^2=r^2[/itex] and [itex]x[/itex] and [itex]y[/itex] each go from 0 to 1, so wouldn't [itex]r[/itex] go from [itex]\sqrt{0^2+0^2}=0[/itex] to [itex]\sqrt{1^2+1^2}=\sqrt{2}[/itex]?
     
  4. Aug 13, 2009 #3

    kuruman

    User Avatar
    Homework Helper
    Gold Member

    I am not sure about the square root of 2; Since at constant z you have a circle in a plane parallel to xy, x and y are related to r by the parametric equations

    x = r cosφ y = r sinφ

    When x = 1, y = 0. Am I missing something?
     
  5. Aug 13, 2009 #4

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Nope, my mistake...just a small brainfart
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Different approach on calculating flux threw area
Loading...