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A surface integral

  1. Mar 17, 2013 #1
    Hi to all,

    1. The problem statement, all variables and given/known data
    Evaluate the surface integral of the vector F=xi+yj+zk over that portion of the surface x=xy+1
    which covers the square 0≤x≤1 , 0≤y≤1 in the xy plane

    2. Relevant equations
    ∫∫F.ndσ
    n=∇g/|∇g|

    maybe transformation to the volume integral

    3. The attempt at a solution

    g(x,y,z)=xy+1-z
    n=∇g/|∇g|=(yi+xj-k)/√(y2+x2+1)

    Plugging into integral,
    i finally got

    ∫∫((xy-1)/(√(y2+x2+1))) dxdy

    both x and y are from 0 to 1.
    But i could not take this integral without help of a computer.

    Since this is from a book's ordinary question i don't think it needs such a treatment.

    I think i could transform the surface integral into a volume integral but there is not a well defined volume that can be used.

    So, i stuck at this point.

    Thanks.
     
  2. jcsd
  3. Mar 17, 2013 #2
    ok, drawing the region, transforming into volume integral and subtracting three additional surface integrals; i finally found 3/4 with only pencil and paper.

    thanks anyway.
     
  4. Mar 18, 2013 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    From what you write below, you mean z= xy+ 1.

    Do you mean using some version of Stoke's theorem? Those all require a closed surface which you do not have here.

    You are confusing the integral of a scalar valued function with the integral of a vector valued function. What you are doing uses some parts of both!

    Instead of the surface area integral you should use the vector surface differential.
    Some texts write that as [tex]\vec{u}dS[/tex] where u is a unit vector perpendicular to the surface, say the gradient, that you calculated for dS. But that require multiplying by the length of grad s in dS and diving by it in u. And they will cancel anyway.

    Instead, write the surface as the xy+1- z= 0 or, even simpler, xy- z= -1, which are just different ways of writing the same surface. Now, grad(xy- z)= <y, x, -1> and the "vector differential of surface area is [itex]d\vec{S}= <y, x, -1>dx dy[/itex] (because only x and y appear in that normal vector, it is simplest to choose x and y as parameters).

    Now, the vector function to be integrated is [itex]\vec{F}= xi+ yj+ zk= xi+ yj+ (xy+ 1)k[/itex] on this surface where I have written z in terms of the parameters x and y.

    So [itex]\vec{F}\cdot d\vec{S}= <x, y, (xy+ 1)>\cdot <y, x, -1>dxdy= xy+ xy- xy- 1= xy- 1[/itex]

    The integral you want is [itex]\int_{x=0}^1\int_{y= 0}^1 xy- 1 dydx[/itex]

    But, you did not specify an orientation for the surface here! You cannot do a vector surface integral without specifying an orientation for the surface. I, since I use <y, x, -1> as normal vector, am assuming downward (the z component is negative) orientation. If the problem specifies an upward or orientation, use <-y, -x, 1> which would just multiply the final result by -1.
     
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