double intergrals

[Kamataat]

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

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

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

Thanks in advance for your help!!!

[quasar987]

Is it assumed that f(P_i) is constant everywhere for some D_i, so that f(P_i) won't have different values depending on where in D_i we choose P_i?

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.

[TheDestroyer]

As you know, if we want to get the area of the surface D, we just perform the integral:

\int\int dx dy

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

\int\int\int dx dy dz

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:

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

and doing a first step will give:

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

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,