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Difference between these two surface integrals?

  1. May 24, 2010 #1
    In my Calculus book, in the chapter that introduces multiple integration, it has a chapter on integrals that calculate the surface area of a function in space. They define the integral to be..

    [tex] \int \int dS = \int \int \sqrt{1+\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2}dA [/tex]

    However, one chapter later they have another chapter entitled "Surface Integrals" where they define the surface area of a function in space to be..

    [tex] \int \int f(x,y,z) dS = \int \int f(x,y,g(x,y)) \sqrt{1+\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2}dA [/tex]

    What is different between these two integrals? They both say they calculate surface area.

    EDIT:

    From what I can gather, the first one is for an area right under the function in space, and the second one is for a region other than the base under the curve. Is this correct?

    Thanks!
    ~Matt
     
    Last edited: May 24, 2010
  2. jcsd
  3. May 24, 2010 #2
    The second one looks like it's integrating the value of a function over a surface. It's hard to answer since we don't know what all the variables mean, or even what space we're working in (though I assume 2 or 3D space).

    Wouldn't be able to take a picture of the textbook would you? Or if it's the Stewart's essential transcendental crocodiles or whatever that book is, I could go get it quick.
    [edit]
    Also, functions don't really have areas.
    I know we've all been taught that the integral of y(x)dx is the area under a curve...
    but you can't hold onto that way of thinking forever (even though you can generalize it somewhat for a while).

    Shapes in space have areas associated with them.
    Functions assign a number (or something like a number) to each point in space.
    Well anyway, in my first vec calc stuff, I found it easier to think of integrals as sums.
     
    Last edited: May 24, 2010
  4. May 24, 2010 #3
    Cant take a picture at the moment, but I think you may be right. I think the first one Calculates the surface area of the actual curve in space, and the second one calculates the function values over another surface.
     
  5. May 24, 2010 #4
    What's this g(x,y) thing?

    Oh. The surface is z=g(x,y)?
    Not sure why the derivatives of 'f' appear in the sqrt, but it's early in the morning Dx
     
  6. May 26, 2010 #5

    LCKurtz

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    That is the area of as surface parameterized as z = f(x,y). The double integral is taken over the (x,y) domain of the function f, and the value of the integral is the surface area of the surface given by z = f(x,y).

    Are you sure that one is copied right? More typical would be something like

    [tex] \int \int \delta(x,y,z) dS = \int \int \delta(x,y,f(x,y)) \sqrt{1+\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2}dA [/tex]

    This would represent the mass of the same surface where its variable area density is [itex]\delta(x,y,z)[/itex]. If the density is 1 it gives the same formula for surface area as above.

    It isn't the area "right under" the graph. It is the surface area of the surface itself.
     
  7. May 27, 2010 #6
    depends what your doing. are you looking at a single surface of a function?, or a function with a boundary over a surface?
     
  8. May 27, 2010 #7
    LCKurtz, what you say makes sense, and there are examples that find the mass of a surface with a variable density. I did copy it down incorrectly. I understand now. Thanks :D
     
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