# Calculating Volume of Cylinder with Double Integrals

• Kamataat
In summary, the book defines a function, f, that is constant over the area D, and uses the formula V_i=f(P_i)\Delta S_i to find the volume V of an object that is above an area D_i. The book then goes on to explain that if you want to find the volume of an object that is not above an area, you will need to use the triple integral.
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 $n\to\infty$.

Kamataat said:
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.

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,

## 1. How do you calculate the volume of a cylinder using double integrals?

To calculate the volume of a cylinder using double integrals, you will need to integrate the appropriate function over the specified limits. The formula for calculating the volume of a cylinder using double integrals is ∫∫R f(x,y) dA, where R is the region of integration, f(x,y) is the function, and dA is the differential area element. This involves performing two integrals, one for the height of the cylinder and one for the circular base.

## 2. What is the difference between using single and double integrals to calculate the volume of a cylinder?

The main difference between using single and double integrals to calculate the volume of a cylinder is that single integrals can only be used for simple, two-dimensional shapes, while double integrals can be used for more complex, three-dimensional shapes. In the case of a cylinder, a single integral can be used to calculate the volume if the height of the cylinder is a constant, but a double integral is needed if the height varies.

## 3. What are the limits of integration when calculating the volume of a cylinder using double integrals?

The limits of integration when calculating the volume of a cylinder using double integrals will depend on the dimensions of the cylinder. For the height of the cylinder, the limits will typically be from 0 to the height of the cylinder. For the circular base, the limits will depend on the radius of the cylinder and the orientation of the cylinder (vertical or horizontal).

## 4. Can double integrals be used to calculate the volume of any three-dimensional shape?

Yes, double integrals can be used to calculate the volume of any three-dimensional shape as long as the shape can be described by a function of two variables (x and y). However, the limits of integration and the complexity of the integrand may vary depending on the shape.

## 5. What are some practical applications of using double integrals to calculate volume?

Double integrals can be used to calculate the volume of complex shapes in fields such as physics, engineering, and geometry. For example, they can be used to calculate the volume of a solid object or the amount of fluid flowing through a pipe. They are also commonly used in computer graphics to create three-dimensional models of objects or environments.

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