# 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,