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:
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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