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!

Surface Integrals

  1. Apr 23, 2007 #1
    1. The problem statement, all variables and given/known data

    Evaluate ∫∫S √(1 + x^2 + y^2) dS where S is the helicoid: r(u, v) = ucos(v)i + usin(v)j + vk, with 0 ≤ u ≤ 4, 0 ≤ v ≤ 3π

    3. The attempt at a solution

    What I tried to do was say x=ucos(v) and y=usin(v), then I plugged those into the sqrt(1+x^2+y^2) eq, which I ended up simplifying to sqrt(1+u) somehow. I then took the cross product of the surface eq differentiated (respect to u) with eq (respect to v). I found the length of that, which I somehow got to be sqrt(2).

    I then said that the intergal was ∫∫sqrt(1+u)*sqrt(2)du*dv
    where the limits of integration are the limits of u and v given above.
    Well this is wrong. Hah!

    Any help would be appreciated.
  2. jcsd
  3. Apr 24, 2007 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    I'm not happy with the "somehow"!! That's where you went wrong. The cross product you speak of (also called the 'fundamental vector product') is [itex]sin(v)\vec{i}- cos(v)\vec{j}+ u\vec{k}[/itex]
    (did you forget the "u" multiplying -cos(v) and sin(v) in the derivative with respect to v?) and that has length [itex]\sqrt{1+ u^2}[/itex]. Thus your integral is
    [tex]\int_{u= 0}^4\int_{v=0}^{2\pi}(1+ u^2)dudv[/tex]

    Last edited: Apr 24, 2007
  4. Apr 24, 2007 #3
    Thanks for the reply.
    I still cannot seem to get the correct answer. I solved the double integral you provided and was incorrect, then changed the limit on the integral with respect to u to 3pi like the limits given, but still no luck. Any suggestions on what I'm doing wrong? Thanks.
    Last edited: Apr 24, 2007
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Surface Integrals
  1. Surface integral (Replies: 6)

  2. Surface integrals (Replies: 10)