Application of integral: the volume formula

In arithmetic we learn the formula to calculate the volume of a  a solid such as a cube. The problem is we have to memorize the formulas for every figure. The definite integral allows us to find a general formula  for the volume of any solid.
Let's consider a cubic solid limited by two parallel planes perpendicular to the x-axis. At any point x, any plan perpendicular to the x-axis is called a cross section.


In order to find the volume of this solid, we are going to divide it in different slices of widths 𝝙x₁, 𝝙x₂, 𝝙x₃ 𝝙xn. This leads to divide the interval [a,b] into different sub-intervals of lengths 𝝙x₁, 𝝙x₂, 𝝙x₃ 𝝙xn. Let's A₁, A₂, A₃... An be the different cross-sections or base of the cubic slices. The volume of the cube is equal to the sum of the volume of the different slices.

V = A₁𝝙x₁ + A₂𝝙x₂ + A₃𝝙x₃ + An.𝝙xn.

There is a way to write this sum simpler:

Let's divide each slice into more slices. We now have thinner slices. Let's continue to divide each slice into more slices. The slices become thinner and thinner. At each time we get a better approximation. We can notice also that each time we divide each slice the width becomes smaller and smaller. It means that 𝝙x approaches zero. When this is the case v approaches a certain value. This value represents the limit of the sum and is the value of V. We can write;

This value represents the definite integral of the function A(x) when Δx approaches zero. We can finally write in the integral form:

Application

Let's demonstrate that the volume of a pyramid is equal to one third the area of its base by its height.
If a is the length of the sides of the base and h the height V = 1/3 a²h



Since the cross section is perpendicular to the y-axis we have to integrate in respect to y. Then the volume of the pyramid can be calculated by

The cross-section A(y) varies from o to h therefore the volume V becomes

A(y) is a square of sides b. Therefore its area is b². We have A(y) = b².

Let's calculate b. To do that, let's isolate from the figure above the triangle with height h

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Area between two curves

Objective: Find the area between two curves

Method

Let's consider 2 functions f and g. We want to find the area limited by the curves of these functions and the verticals x = a and x = b.

.
The area limited by the the curves of f and g  and the two verticals passing respectively by x = a and x = b can be found by subtracting the area under the curve of g from the area under the curve of g.

Let A be the area between the 2 curves we can write:
A = Area under f-Area under g
The area under f and limited by the verticals passing by a and b and the x-axis is given by

The area under g limited by the verticals passing by a and b and the x-axis is given by

Let's substitute area under f area under g in A

According to this result, we can make the following statement:

Let f and g be two continuous functions on the interval [a,b] and f(x)≥ g(x) for all values of x in this interval, then the area between the the curves of and g is given by

Example

Solution

                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                             
Practice 

Sketch the region enclosed by the curves of y = x²and y = x + 3. Find the area. 

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Applications of integral: area between a curve and the x- axis

Objective: Find the area between the curve of a function f and the x-axis over an interval [a b].

Definition

The area between the curve of a function and the x-axis bounded by the verticals x = a and x = b is given by

The area is positive if it is located above the x-axis and negative if it is below.



The areas A(2), A(4) are negative. The areas A(1), A(3) and A(5) are positive. The total area is the sum of all the areas: positive and negative.

Example 



Solution

Practice

  

 

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