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      Calculus II

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Calculus II: Volume of solid of revolution



In the area calculus section, we used definite integral with two kinds of cut: the vertical slice and the horizontal slice.

In this section, we will use them determine the volume of a solid of revolution.

If we rotate a solid object about a rotation axis, each point of this solid makes a circular orbit around this axis.
To define the related generated volume we will use two method: the disk method, and the cylindrical method.

In practise, the disk method is used with slices perpendicular to the rotation axis and the cylindrical method is used with slices parallel to the rotation axis.

1.Volume of solid of revolution
Disk method

1.1. Rotation axis is x-axis




We want to determine the volume generated by the rotation of the area delimited by the function f(x) and the rotation x-axis, between the two bounds x = a and x = b.

Because we are using the disk method, let's cut slices perpenticular to the rotation axis x-axis.

At the point x, the slice has f(x) as radius and dx as thickness.

So the area of the elementary disk at the point x is
disk(x) = π[f(x)]2.

The related elementary volume dV is

dV(x) = disk(x) dx = π[f(x)]2 dx .

The total volume generated by the complete rotation is
  b  
V =   π [f(x)]2 dx
  a  



1.2. Shifted rotation axis

If the rotation axis is shifted by y = c around the x-axis, we will have f(x) - c as the radius at the point x, so


  b  
V =   π [f(x) - c]2 dx
  a  



1.3. General case

Let f and g two functions continuous on the interval [a,b]. If within the whole interval [a, b], f(x) > g(x) > c, then the volume of the solid of revolution generated by the rotation of the area bounded by the curves of the two functions between x = a and x = b about the c axis is:
  x = b  
V =   π[(f(x) - c)2 - (g(x) - c)2]dx
  x = a  


To determine the related volume, we will use Riemann sum integral with vertical slices cut.

Let subdivise the interval [a, b] in n equal sub-intervals of width Δx1, Δx2, Δx3, ... Δxn and take the upper bound of each sub-interval as the representative.

In each subinterval, we construct a rectangle of width equal to Δxi and height f(xi) - g(xi).

A complete rotation about c axis of all the n rectangles generates a n washers. The volume of the ith washer is:

Vi = π[(f(xi) - c)2 - (g(xi) - c)2] Δxi

Therefore the sum of all these washers gives the related volume.
    i = n  
V = lim   Σ   Vi
n →+∞  i = 1  
  x = b  
V =   π[(f(x) - c)2 - (g(x) - c)2]dx
  x = a  





1.4. Examples

1.4.1. Example 1

Generated volume by the rotation of the area bounded by the two curves of f(x) = x2 and g(x) = 0 between 0 and x = 1 around x = c = 0

We have: g(x) = c = 0, so
  x = 3  
V =   π[(f(x))2]dx = πx4dx
  x = 0  
=
  1  
[πx5/5]   = π/5  
  0  



1.4.2. Example 2

Generated volume by the rotation of the area bounded by the two curves of f(x) = x2 and g(x) = 0 between 0 and x = 1 around x = c = 1

We have: g(x) = 0, so
  x = 1  
V =   π[(c - g(x))2 - (c - f(x))2]dx
  x = 0  
=
  x = 1  
  π[(1 - 0)2 - (1 - f(x))2]dx
  x = 0  
=
  x = 1  
  π[1 - (1 - x2)2]dx
  x = 0  
=
  1  
π[2x3/3 - x5/5]   = 7π/15  
  0  



1.4.3. Example 3

Generated volume by the rotation of the area bounded by the two curves of f(y) = √y and g(y) = 1 between 0 and 1 = 1 around y = c = 10

We have:

f(y) = x = √ y;
g(y) = 0, so
  y = 1  
V =   π[(f(y) - c)2 - ( g(y) - c)2]dy
  y = 0  
=
  y = 1  
V =   π[(√y - 0)2 - ( 0 - 0)2]dy
  y = 0  
=
  y = 1  
V =   π[y]dy
  y = 0  
=
  1  
π [y2/2]   = π/2  
  0  



1.4.4. Example 4

f(x) = x2 between 0 and x = 1 around c = y = 1 Generated volume by the rotation of the area bounded by the two curves of f(y) = √y and g(y) = 1 between 0 and 1 = 1 around y = c = 1

We have:

f(y) = x = √ y;
g(y) = c = 1, so
  y = 1  
V =   π[(c - f(y) )2 - (c - g(y))2]dy
  y = 0  
=
  y = 1  
V =   π[(1 - √y)2 - ( 0 - 0)2]dy
  y = 0  
=
  y = 1  
V =   π[1 - 2√y + y]dy
  y = 0  
=
  1  
π [y - (4/3)y3/2 + y2/2]   = π/6  
  0  



2.Volume of solid of revolution
Cylindrical method

Let f and g two continuous functions on the interval [a,b]. If within the whole interval [a, b], f(x) >= g(x), then the volume of the solid of revolution generated by the rotation of the area bounded by the curves of the two functions between x = a and x = b about the c axis with a >= c is:
  x = b  
V =   2π(x - c)(f(x) - g(x)) dx
  x = a  



We want to determine the volume generated by the rotation of the area delimited by the functions f(x) and g(x) about c-axis, between the two bounds x = a and x = b.

Because we are using the cylinder method, let's cut shells parallel to the rotation axis c-axis.

At the point xi, the shell has xi - c as radius (c<0) and Δxi as thickness.

So the area of the elementary rectangle side at the point xi is
shell(xi) = height x Δxi = [f(xi) - g(xi)]Δxi.

The elementary volume dV of this shell

dVi = dV(xi) = shell(x) . 2π[xi - c] = 2π[xi - c] [f(xi) - g(xi)]Δxi.

The total volume generated by the complete rotation is
    i = n  
V = lim   Σ   Vi
n →+∞  i = 1  
=
  x = b  
  2π[x - c] [f(x) - g(x)] dx
  x = a  



5. Exercises




  


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