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Double intergrals

  1. Apr 3, 2005 #1
    To calculate the volume of a cylinder that has as its bottom (or top) end the area [itex]D[/itex] in the xy-plane, we divide [itex]D[/itex] into [itex]n[/itex] smaller areas [itex]D_i (i=1;...;n)[/itex]. The function [itex]f(x,y)[/itex] is defined at every point [itex]P(x,y)[/itex] of [itex]D[/itex], in short [itex]f(P)[/itex]. So, to find the volume [itex]V_i[/itex] that is above an area [itex]D_i[/itex], we use the formula [itex]V_i=f(P_i)\Delta S_i[/itex], where [itex]f(P_i)[/itex] is the height of the cylinder above [itex]D_i[/itex] and [itex]\Delta S_i[/itex] is the area of [itex]D_i[/itex]. Now to get the volume of the space above [itex]D[/itex], we calculate [itex]V=\sum_{i=1}^n f(P_i)}\Delta S_i[/itex].

    Now this is what I don't understand: How can one get the volume above [itex]D_i[/itex] from just [itex]P_i[/itex]? Is it assumed that [itex]f(P_i)[/itex] is constant everywhere for some [itex]D_i[/itex], so that [itex]f(P_i)[/itex] won't have different values depending on where in [itex]D_i[/itex] we choose [itex]P_i[/itex]?

    In the definition of the double integral they (the book) say that [itex]n \to \infty[/itex], which is understandable. However, before ever getting to double intergrals, they give the formula for [itex]V[/itex], w/o [itex]n\to\infty [/itex].

    Thanks in advance for your help!!!
  2. jcsd
  3. Apr 3, 2005 #2


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    Well, if the volume you are trying to compute is that of an ordinary cylinder, then f not only IS constant over each Di, f is constant over D, and it is the constant function f(x) = h where h is the height of the cylinder.

    So the sum w/o n --> infty does indeed give the exact value of the volume.

    However, it the volume you were trying to compute was that of a truncated cylinder or any other irregular form, then the sum w/o n --> infty would only be an approximation of the volume and an infinity of Di would have been required to get the exact value.
  4. Apr 3, 2005 #3
    As you know, if we want to get the area of the surface D, we just perform the integral:

    [tex]\int\int dx dy[/tex]

    And in the same way, if we want to find the volume using the Triple Integrals we are going to perform the integral:

    [tex]\int\int\int dx dy dz[/tex]

    So when you release the triple integral when you define the z from the surface xy-plane to any function you want f(x,y) you get:

    [tex]\int\int\int dx dy dz = \int_{a}^b \int_{y_1(x)}^{y_2(x)} \int_{0}^{f(x,y)} dzdydx[/tex]

    and doing a first step will give:

    [tex]\int_{a}^b \int_{y_1(x)}^{y_2(x)} f(x,y) dy dx[/tex]

    Here you can see that we got the double integral you talked about, so we can consider the double integral for a volume a second step for the triple integral which begins from z=0 to z=f(x,y),

    I hope my explanation let you understand, if you have just started with integrals be patient, later you will get everything clear with triple integrals,
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