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 (Stuck halfway)

  1. Nov 19, 2014 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    ∫∫D F((r(u,v))⋅(ru x rv) dA

    3. The attempt at a solution


    I got stuck after finding the above, at where the double integrals are. :(

    May I know how do I find the limits of this? (I always have trouble finding the limits to sub into the integrals. ><)

    Thank you! :)
  2. jcsd
  3. Nov 19, 2014 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You have chosen u = x and v = y. This is just like a problem you posted earlier. Draw a picture of the slanted plane in the first octant and use the triangle in the xy plane for your limits.

    [Edit, added] One thing that may be causing you confusion is the use of ##u,v## parameters when they are unnecessary. You usually use those when there is a reason to change coordinate systems. In a problem like this, ##x## and ##y## are natural parameters and there is no need to use ##u## and ##v##. Just let ##x=x,~y=y,~z= 2-2x-y## and$$r(x,y)=\langle x,y,2-2x-y\rangle$$and look in the ##xy## plane for the limits.
    Last edited: Nov 19, 2014
  4. Nov 19, 2014 #3
    Oh dear! Haha. I didn't realise that this question is similar to the one I have asked earlier. I am currently focusing on past year papers as my Maths exam is just 2 days away. Sigh~ I find that I always have a problem visualising the domain... because I didn't know the domain was a triangle until you told me above and after doing some look-up on Google. >< Sheesh. :(

    I will take note of what you have said and try again. :) I will post back here again!
  5. Nov 19, 2014 #4


    User Avatar
    Homework Helper

    A linear equation in three variables represents a plane.

    The given equation ##z = 2 - 2x - y## is a plane. Another example would be ##x + y - z = 1 \Rightarrow z = x + y - 1##.

    If you want to have an easier time visualizing, then you can easily solve for the intersections. Consider the points ##(x, 0, 0), (0, y, 0), (0, 0, z)## for the plane equation in your problem.

    Choosing the point ##(x, 0, 0)##, you would get ##z = 2 - 2x - y \Rightarrow x = 1##. In a similar fashion you would get ##y = 2## and ##z = 2##. Plotting these solutions on a 3D graph and connecting the dots will give you a nice visual of the plane.

    This will also help you visualize the projection of the plane onto the x-y plane. Drawing this projection in the x-y plane will give you your limits for integrals.
  6. Nov 20, 2014 #5
    I see. Thank you. My teacher changes them to r(u,v) for the chapter of surface integrals, hence I would always change it to r(u,v), though I don't know why my teacher uses the u,v coordinate system. I think I should ask him some day. Haha.

    Ah yes, I got the triangle in the xy-plane as shown below:


    Just wondering, how can I be very sure that I have placed the limits on my ∫ ∫ correctly? My integral equation is below:


    Thank you! :)
    Last edited: Nov 20, 2014
  7. Nov 20, 2014 #6
    Thanks, Zondrina! Your method is really effective! :DI think I can now at least find the intersection points of the lines on the x-y-z axes, though I feel it can be quite difficult to plot/sketch/draw a 3D figure on a 2D piece of paper.
  8. Nov 20, 2014 #7


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Your limits are OK but I think you have u and v exchanged in the integrand. Another reason not to use u and v unnecessarily.
  9. Nov 20, 2014 #8
    I tried integrating the above and I got the same answer as my teacher, who did it in another way:


    Hence, I think both methods work (or is this my method a 'fluke'? Haha.) Thanks. :)
  10. Nov 20, 2014 #9


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    No you didn't. In post #7 you wrote the integrand as ##u-v+2##, which has the ##u## and ##v## switched as I said. Below you have the correct integrand.

  11. Nov 20, 2014 #10
    I see. Haha. I think I misinterpreted your intended message. Sorry. Thanks for all your help! :D
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Surface Integrals (Stuck halfway)
  1. Stuck with an Integral (Replies: 10)

  2. Stuck with an integral (Replies: 5)