Double intergrals

1. Apr 3, 2005

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 $n\to\infty$.

2. Apr 3, 2005

quasar987

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.

3. Apr 3, 2005

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,