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What is dS?

  1. Apr 26, 2012 #1
    1. The problem statement, all variables and given/known data

    My question is not specific to any particular problem but is rather of a conceptual nature. In my advanced calculus class this semester, the notation ds appears often, for line integrals, surface integrals, and arc length. In all honestly, I don't really understand what ds is and, as such, have had a lot of trouble solving problems involving it. Furthermore, while I know all of the formulas, I don't really understand how or why they work. Could somebody help me out on this one?

    2. Relevant equations

    arc length:
    [tex]ds = \sqrt{ \left( \frac{dy}{dx} \right) ^2 + 1} [/tex]

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

    3. The attempt at a solution
    As far as I can tell, the element ds seems always to reduce the dimension of the function by 1. Perhaps it represents a projection onto that dimension? Obviously, if it's used to calculate surface area of solids and arc length of 2D functions, something like that must be going on, right? Am I on the right track?

    Thanks,
    Trapezoid
     
  2. jcsd
  3. Apr 26, 2012 #2
  4. Apr 26, 2012 #3
    As Dustinsfl said, it's just arc length

    What does this mean you might ask?

    Take some curve and then zoom in to the infinitesimal level, how long is each little infinitesimal length of curve?
    Through pythagoras we see that the length of the little segment should be
    [itex]ds=\sqrt{dx^2 + dy^2}[/itex]
    This is a bit awkward to work with however, so let us divide through by [itex]dx[/itex] to get
    [itex]ds=\sqrt{1+\frac{dy}{dx}^2}dx[/itex]
    Adding up the length of all these little segments, which is just the integral over ds, will then give you the arc length
     
  5. Apr 26, 2012 #4

    Mark44

    Staff: Mentor

    To be clearer, I would write the above as
    $$ds=\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx $$
     
  6. Apr 26, 2012 #5

    sharks

    User Avatar
    Gold Member

    To find the arc length of the curve C:
    [tex]\int_C \,ds[/tex]
     
  7. Apr 26, 2012 #6
    Thanks to everyone for your quick replies and kudoes to genericusrname - that explanation really helped. I still, however, have a bit of trouble seeing how it relates to surface integrals.

    An infinitesimally small portion of the surface of a solid is represented by [itex]ds = \sqrt{dx^2 + dy^2 + dz^2}[/itex] but dividing by dz looks like [itex]ds = \sqrt{\left( \frac{\partial x}{\partial z} \right) ^2 + \left( \frac{\partial y}{\partial z} \right) ^2 + 1} \: dz[/itex] which doesn't look right. If z is a function of x and y, then perhaps we need to divide by the element dA but, again, I don't see how that would work.

    Any tips? Have I completely misunderstood? :tongue2:
     
  8. Apr 26, 2012 #7

    sharks

    User Avatar
    Gold Member

    Try to multiply the dz into the square root sign and you'll see. :smile:
     
  9. Apr 26, 2012 #8

    Mark44

    Staff: Mentor

    I'm guessing that x, y, and z are independent variables of, say, t. If so, then the above should look like
    [tex] ds = \sqrt{\left( \frac{dx}{dt} \right) ^2 + \left( \frac{dy}{dt} \right) ^2 + \left( \frac{dz}{dt} \right) ^2} dt[/tex]
     
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