1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: 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

    maybe transformation to the volume integral

    3. The attempt at a solution


    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.

  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


    User Avatar
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted