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Basic question on double integrals

  1. Apr 2, 2015 #1
    I am just starting to learn double and triple integrals.
    Say:
    I = ∫ ∫ dx dy
    This should give the area within two curves right?
    What will the following integral give? Will it give volume? I am finding questions where it gives (for instance) mass in a given shape
    I = ∫ ∫ f(x,y) dx dy
     
    Last edited: Apr 2, 2015
  2. jcsd
  3. Apr 2, 2015 #2

    MarcusAgrippa

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    It depends on how you interpret the symbols.

    Your integral could be interpreted as a volume provided that you interpret f(x,y) as a function that defines a surface in 3 dimensional space. So, put z = f(x,y). This defines a surface which projects nicely onto the x-y plane (that is, it never turns vertical). Subdivide the x-y plane into small rectangular segments of sides dx and dy. The segment at position (x,y) has area dx dy , and the height of the surface above the point (x,y) is f(x,y). So the product f(x,y) dx dy is approximately the volume of the rectangular column with base dx dy that reaches up to the surface z = f(x,y). Integration is the process of adding these volumes in the limit as dx and dy tend to zero, and gives the (signed) volume of the vertical solid column defined by the boundary curve on the x-y plane that defines the limits of the integral with vertical walls reaching up to the surface z = f(x,y).

    An alternative interpretation is the following. Suppose you have a thin sheet of material (a lamina) with density (more precisely, with mass per unit area) given by f(x,y). Then a small rectangle of material with sides dx and dy will have area dx dy and therefore mass f(x,y) dx dy. Add up the masses of these small rectangles to get the mass of the lamina. In this interpretation, the integral does not describe a 3-dimensional object, but a two dimensional one.

    Remember that maths is form without content. The moment you interpret the symbols in a way that associates them with something physical, you are adding content to the maths. A single piece of mathematics is often capable of very different physical or geometrical interpretations. Different physical systems can be described by the same mathematical model.
     
    Last edited: Apr 2, 2015
  4. Apr 2, 2015 #3
    No. the integral depends on the meaning of the function f(x,,y). If it were mass density for example the integral would be the total mass of the area encompassed by the curve. If it where Pressure it would be the total force on the area. So it is the summation of a quantity whose value depends on its position with the closed curve times the elemental area dxdy.

    If f(x,y)⋅dx⋅dy =dF(x,y)

    then ∫∫ f(x,y) dx dy = F(x,y)
     
  5. Apr 2, 2015 #4
    One more question.
    Take this problem:
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    ?temp_hash=64ec2ddeebdb5d4738258166325866f7.png


    I don't get how the upper limit for x is (1-y). I get that the limits for Y is 0-1. if I get the limits for x to be from 0-1, will the surface be a square ?
     

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  6. Apr 2, 2015 #5

    SteamKing

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    Along the line drawn from (0, 1) to (1, 0), the equation x + y = 1 must be satisfied. By re-arranging this equation, x = 1 - y, which forms the upper limit when integrating along the x-axis.
     
  7. Apr 3, 2015 #6

    HallsofIvy

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    If you decide to integrate in the order "dydx" so that dx is the "outer" integral, then the limits of integration must be constants so that the integral is a constant. The line x+ y= 1 cuts the x-axis at x= 1 so those limits must be 0 to 1. Now, for each x, y must range from the x-axis, y= 0, up to the boundary line x+ y= 1. That is, y must range from y= 0 to y= 1- x.
     
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