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Surface integrals

  1. Jul 22, 2012 #1
    1. The problem statement, all variables and given/known data
    Evaluate ∫∫F.nds where F=2yi-zj+x^2k and S is the surface of the parabolic cylinder y^2=8x in the first octant bounded by the line y=4, z=6


    2. Relevant equations

    We were told that the projection is supposed to be taken in the yz plane but how?? and i have a feeling that i did something wrong with finding n, are we supposed to use y^2=8x for that?
     
  2. jcsd
  3. Jul 22, 2012 #2

    LCKurtz

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    I would use the parameterization ##\vec R =\langle \frac{y^2}8,y,z\rangle## and use the formula$$
    \pm\int_0^4\int_0^6 \vec F \cdot \vec R_y\times \vec R_z\, dydz$$ with the sign chosen depending on whether the direction of ##\vec R_y\times \vec R_z## agrees with the orientation, which you didn't give.
     
  4. Jul 23, 2012 #3

    sharks

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    OK, so in other words, you're being asked to find the flux of the field ##\vec F=2y\vec i-z \vec j+x^2\vec k## across the surface S.

    First, you should draw the surface and the given 2 planes in the 1st octant so you can better understand the limits and the projection. Projecting the surface S onto the yz-plane will give you an area as shown in the attached figure.

    If you use the method as described by LCKurtz, you'll probably find it easier, since it's apparently the standardized method for solving this kind of problem. However, i was taught to use another method, and since it's stuck in my mind, i'm going to use it here and maybe it'll help you to understand better, coupled with whatever method you might already know/use.
    $$\phi (x,y,z)=-8x+y^2
    \\\nabla \vec \phi=-8\vec i +2y\vec j
    \\\hat n =\frac{-8\vec i +2y\vec j}{\sqrt{64+4y^2}}
    \\Flux = \int^6_0 \int^4_0 \frac{-8y-yz}{\sqrt{16+y^2}}\,.dydz
    $$Solving this double integral should give you the final answer. You could also have projected the surface S onto the xz-plane.
     

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    Last edited: Jul 23, 2012
  5. Jul 23, 2012 #4

    LCKurtz

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    No, it won't give you the correct answer. You have calculated ##\vec F\cdot \hat n## correctly except for not knowing the orientation, but the surface element on the surface is not ##dS=dydz##.
     
  6. Jul 23, 2012 #5

    sharks

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    [color="blue]Mod note: deleted full solution[/color]

    In the recent past and yet still now, i've struggled to fully grasp and understand the method that you use for this type of problem, LCKurtz. But unfortunately, i haven't found a good set of notes on the matter to rebuild what i know from the ground up, so i always revert back to what i learned in class. I don't quite understand why my lecturer would put us through this apparently unorthodox method, as i've seen the formula that you use more often while researching this topic.
     
    Last edited by a moderator: Jul 23, 2012
  7. Jul 23, 2012 #6

    LCKurtz

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    Maybe looking at posts #12 and #13 in this thread

    https://www.physicsforums.com/showthread.php?t=611873

    will help you.
     
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