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      Calculus II

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Calculus II: Volume of solid of revolution



We have seen that the length of an arc describes by a function f(x) is:

dL = [1 + (f '(x))2] 1/2 dx

We use this result to determine the lateral area (curved surface) generated by the revolution of a curve around an axis.



1.Lateral area generated by a curve




Let f(x) a function continuous on the interval [a, b], and f '(x) its continuous derivative on ]a, b[.

We want to determine the area generated, between x = a and x = b, by the revolution of the curve of f(x) around the x-axis.

Let divide the interval [a, b] in n equal sub-intervals of width Δx2, Δx2, ..., Δxi-1, Δxi, ..., Δxn.

For each point x = xi corresponds its image f(xi) and the elementary arc length ΔLi = [1 + (Δf(xi)/Δxi)2]1/2, where
Δf(xi) = f(xi) - f(xi-1), and
Δxi = xi - xi-1.

By a revolution around the x-axis the elementary arc length ΔLi generates an elementary lateral area :

ΔAi = ΔLi . 2π f(xi)

The total area A is the sum of the contributions of all the elementary Ai; so

    i = n  
A = lim   Σ   Ai
n →+∞  i = 1  

=
    i = n  
2π lim   Σ   f(xi) [1 + (Δf(xi)/Δxi)2] 1/2
n →+∞  i = 1  

=
  x = b  
  f(x) [1 + (f '(x))2] 1/2 dx
  x = a  



2.Example




Lateral area of a sphere of radius R centered on the origin.

We determine first the lateral area of a half-sphere generated by the curve of:

f(x) = [R2 - x2]1/2.

around the x-axis between 0 and R.

The function is continuous on [-R, R] then on [0, R]. Its derivative f '(x) is continuous on ]0, R[

f '(x) = - x/[R2 - x2]1/2

dL = [1 + (f '(x))2]1/2 dx =
= R/[R2 - x2]1/2 dx

Therefore

  x = R  
A = 2π   [R2 - x2] 1/2 R/[R2 - x2] 1/2 dx
  x = 0  

=
  x = R  
2π R   dx
  x = 0  

= 2π R2

Hence

The lateral area of a sphere of radius R is 2 x A = 4π R2



5. Exercises




  


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