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].(adsbygoogle = window.adsbygoogle || []).push({});

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!!!

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# Double intergrals

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