# Explain: What Double Integral represents and write its two applications?

Aug 11, 2018

$V = {\int}_{x = 0}^{x = b} \left({\int}_{x = 0}^{x = b} b \cdot \mathrm{dx}\right) \cdot \mathrm{dz}$

#### Explanation:

Double Integral.

Let us consider determining the volume of a cube with sides b, which is the simplest solid for which the answer is known in advance.
This has two stages.
First stage,
Determining the area of the base.
Here, the base can be imagined as collection of lines with length equal to b in a plane over a length of b.

Let the variable y, denote the length of a line at any distance x, from a selected origin,
Here, y = b

The area enclosed by the elementary strip of length dx, is then given by
dA = y dx
dA = b dx

The area of the base can be visualized as a collection of such elementary strips spanning from the lower limit, x = 0, to the upper limit, x = b
Integrating the area of the strip between the lower limit and the upper limit, we get the area of the entire base.

$A = {\int}_{x = 0}^{x = b} \mathrm{dA}$

$= {\int}_{x = 0}^{x = b} b \cdot \mathrm{dx}$

Since, b is constant throughout the distance from x=0, to x=b

$A = b . x {|}_{0}^{b}$
Evaluating the difference between the limits,

$A = b . \left(b - 0\right)$

$A = {b}^{2}$
Now, having evaluated the area of the base as $A = {b}^{2}$, the first stage is completed.

The cube can be imagined as combination of such bases one above the other vertically upto a height of b.

At any height z, the area of the shape is constant and is equal to A.
The sheet of thickness dz with area A has the volume given by,
$\mathrm{dV} = A \cdot \mathrm{dz}$

The volume of the cube can be visualised as a collection of such elementary sheets spanning from the lower limit, z = 0, to the upper limit, z = b

Integrating the volume of the sheet between the lower limit ad the upper limit, we get the volume of the entire shape.

$V = {\int}_{x = 0}^{x = b} \mathrm{dV}$

$= {\int}_{x = 0}^{x = b} A \cdot \mathrm{dz}$

Since, b is constant throughout the distance from z=0, to z=b,

$V = A . z {|}_{0}^{b}$

$V = A . \left(b - 0\right)$

$V = A . b$

Substituting for A,

$V = {b}^{2} \cdot b$

$V = {b}^{3}$

Consider, again the equation

$V = {\int}_{x = 0}^{x = b} A \cdot \mathrm{dz}$

and

$A = {\int}_{x = 0}^{x = b} b \cdot \mathrm{dx}$

Substituting for A in V, we have;

$V = {\int}_{x = 0}^{x = b} \left({\int}_{x = 0}^{x = b} b \cdot \mathrm{dx}\right) \cdot \mathrm{dz}$

Applications:
With the above explanation, you will be in a better shape to mention many applications, which are more expected achievements.
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