Previously we have already studied the solids in revolution, that is to say in previous post I explained one of the applications that the integral has to calculate the volume of a solid in revolution, in this occasion I am going to explain how to find the area of a solid in revolution.
Definition of surface of revolution
If the graph of a continuous function rotates around a straight line, the resulting surface is a surface of revolution.
Definition of the area of a surface of revolution
Let y = f(x) with continuous derivative in the interval [a,b]. The area S of the surface of revolution formed by rotating the graph of f about a horizontal or vertical axis is:
Area of a surface of revolution
Find the area of the surface formed by rotating the graph of
in the interval [0,1] when rotating around the x-axis.
As a solution, the first thing we are going to do is to graph the cubic function in GeoGebra:
If we take only the portion of the function that goes in x from zero to one and rotate it with respect to the x-axis, we would be left with the following figure:
Therefore the surface area of the solid in revolution when the cubic function rotates around the x-axis is:
Si f(x) = x3
Who is f'(x)?
The derivative of f(x) is:
f'(x) = 3x2
Therefore the area of the surface is:
Then we solve the indefinite integral:
We solve it by the method of substitution or change of variable, where:
u = 1 + 9x4
du = 36x3dx, lo que implica que si despejamos dx:
dx = du/ 36x3
If we clear x3dx:
x3dx = du/36
We make the change of variable:
We solve by the rule of potentiation:
I return the change of variable:
We apply the fundamental theorem of calculus to evaluate from zero to one:
We multiply the result by 2𝜋, since 2𝜋 is the constant left by multiplying by the integral:
(0.5670)(2𝜋)= 3.56
Conclusion and analysis of results
The exercise that was placed as an example to calculate the area of a surface of revolution of the cubic function f(x) = x3 gave a result of 3.56 when rotated with respect to the x-axis.
It is important to emphasize that the integral that had to be solved is a simple integral whose solution lies in applying the method of substitution or change of variable.
Bibliographic Reference
Calculus with Analytic Geometry by Ron Larson, Robert, P. Hostetler and Bruce H. Edwards. Volume I. Eighth Edition. McGraw Hill. Año 2006.
Note: All equations and images were elaborated using the equation insertion tools, and GeoGebra software was also used to elaborate the graphs.
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