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

  1. Feb 8, 2007 #1
    [Can someone help with the following? I am supposed to evaluate the line integral of ∫F∙dr. The curve is oriented counterclockwise as viewed from above. So suppose that F(x,y,z) = (x+y^2)I + (y+z^2)j + (z+x^2)k, and C is the triangle formed by (1,00), (0,1,0), (0,0,1).

    I know that the integral ∫F∙dr over some surface S is equal to ∫∫curlF∙dS and is equal to ∫∫curlF∙n/|n|ds or ∫∫curlF∙ndA, where n is the normal vector. So, curl F is -2zi -2xj -2yk. And since the region has the expression x+y+z = 1, so suppose the region z can be written as (x, y, 1-x-y), and the normal vector to this region has the expression <dz/dx, -dz/dy, 1> or <-1,-1,1>. So wouldn’t curl F dot n be 2z+2x-2y, and since z = 1-x-y, curl F n should be 2-2x-2y+2x-2y? So I am supposed to integrate the integral ∫∫2-4y dydx where 0≤y≤1-x and 0≤x≤1? Can someone point out what I’m doing wrong here? Thanks.
  2. jcsd
  3. Feb 8, 2007 #2


    User Avatar
    Science Advisor
    Homework Helper

    i cant read all that, but couldn't you just check your work by simply parametrizing the triangle and just doing a one variable integral?

    it is especially easy on the vertical and horizontal dies where, on the y axis side say, both dx and dz are zero.
  4. Feb 8, 2007 #3
    But I'm supposed to evaluate it using the double integral, so.
  5. Feb 8, 2007 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    The plane equation is, indeed, x+y+z=1.
    Hence, the normal vector is parallell to (1,1,1)
  6. Feb 8, 2007 #5
    Oh I see where I went wrong! Thanks!!!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook