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Surface Integrals (Stuck halfway)

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

    2. Relevant equations

    ∫∫D F((r(u,v))⋅(ru x rv) dA

    3. The attempt at a solution

    2zxpe0x.jpg

    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

    LCKurtz

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    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

    Zondrina

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    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:

    2ypj2m9.jpg

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

    2n19lxc.jpg

    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

    LCKurtz

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    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:

    2iu3rb6.jpg

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

    LCKurtz

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    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
     
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