Let's consider a function f. positive over an interval [a,b]. Let R be the region bounded by the curve of the function, two verticals x = a and x = b and the x-axis..
A solid of revolution is obtained by revolving the region around the y-axis. Our goal is to determine the volume of this solid. We are going to use the method of cylindrical shells for that purpose.
Method.
Let's divide the region R in small rectangles of width dx and consider one rectangle.
Let's revolve the rectangle around the y-axis. We obtain a cylindrical shell
Let's calculate the volume of this cylindrical shell. The volume of this cylindrical shell is obtained by multiplying the volume of the outer cylinder by the thickness of the cylindrical shell. We can write:
Volume cylindrical shell = Volume outer shell * thickness.
The volume of a cylinder is equal to the surface of the base by the height. Therefore the volume of the outer shell can be written as:
Volume of outer shell = Area of base * height. The base is a circle. Its area is 2𝜋r. Then
Volume outer shell = 2𝜋rh .
Let's substitute in Volume cylindrical shell
Volume cylindrical shell = 2𝜋rhdx (dx is the thickness)
The radius r is variable. We call it x. The height is a function of the radius. We represent it by f(x)
V cylindrical shell = 2𝜋xf(x)dx
Adding the volume of all these small cylindrical shells is going to give the volume of the entire solid of revolution. This is equivalent to integrate the volume over the interval [a,b]. Therefore the volume of the solid of revolution is given by:
Example
y = 1/x y = 0 x = 1 x = 3
Solution
The volume of a cylindrical shell generated by a region around the y-axis is given by
Practice
Find the volume generated by the lines y = 2x-1 y = -2x + 3 and
x = 2 about the y-axis.
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