- #1
evagelos
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Here is another approach to finding the volume :
Let [tex]x_{1}[/tex] be a point in [x.x+Δx] such that :
f([tex]x_{1}[/tex]) = minimum value of f(x) in [x,x+Δx]
Let [tex]x_{2}[/tex] be a point in [x,x+Δx] such that :
f([tex]x_{2}[/tex]) = maximum value of f(x) in [x,x+Δx]
Then,
[tex]\pi.[f(x_{1})]^2.\Delta x\leq\Delta v\leq\pi.[f(x_{2})]^2.\Delta x[/tex]
......OR.........
[tex]\pi.[f(x_{1})]^2\leq\frac{\Delta v}{\Delta x}\leq\pi.[f(x_{2})]^2[/tex]................1Now let :
[tex]\Theta_{1}=\frac{x_{1}-x}{\Delta x}[/tex] ,and
[tex]\Theta_{2}=\frac{x_{2}-x}{\Delta x}[/tex]
.....and (1) becomes:[tex]\pi.[f(x+\Theta_{1}\Delta x)]^2\leq\frac{\Delta v}{\Delta x}\leq\pi.[f(x+\Theta_{2}\Delta x)]^2[/tex].
And as Δx goes to zero,[tex]x+\Theta_{1}\Delta x[/tex] goes to ,x[tex]x+\Theta_{2}\Delta x[/tex] goes to ,x, because [tex]0\leq\Theta_{1}\leq 1[/tex] and
[tex]0\leq\Theta_{2}\leq 1[/tex].You can check that by using the ε-δ definition of a limit.
And since f(x) is continuous in [x,x+Δx],[tex]f(x+\Theta_{1}\Delta x)[/tex] will go to f(x) ,and [tex]f(x+\Theta_{2}\Delta x)[/tex] will also go to ,f(x).
And by the squeezing theorem :
[tex]\frac{dv}{dx} = \pi.[f(x)]^2[/tex] and thus:
[tex]V=\pi\int^{b}_{a}[f(x)]^2dx[/tex]
Let [tex]x_{1}[/tex] be a point in [x.x+Δx] such that :
f([tex]x_{1}[/tex]) = minimum value of f(x) in [x,x+Δx]
Let [tex]x_{2}[/tex] be a point in [x,x+Δx] such that :
f([tex]x_{2}[/tex]) = maximum value of f(x) in [x,x+Δx]
Then,
[tex]\pi.[f(x_{1})]^2.\Delta x\leq\Delta v\leq\pi.[f(x_{2})]^2.\Delta x[/tex]
......OR.........
[tex]\pi.[f(x_{1})]^2\leq\frac{\Delta v}{\Delta x}\leq\pi.[f(x_{2})]^2[/tex]................1Now let :
[tex]\Theta_{1}=\frac{x_{1}-x}{\Delta x}[/tex] ,and
[tex]\Theta_{2}=\frac{x_{2}-x}{\Delta x}[/tex]
.....and (1) becomes:[tex]\pi.[f(x+\Theta_{1}\Delta x)]^2\leq\frac{\Delta v}{\Delta x}\leq\pi.[f(x+\Theta_{2}\Delta x)]^2[/tex].
And as Δx goes to zero,[tex]x+\Theta_{1}\Delta x[/tex] goes to ,x[tex]x+\Theta_{2}\Delta x[/tex] goes to ,x, because [tex]0\leq\Theta_{1}\leq 1[/tex] and
[tex]0\leq\Theta_{2}\leq 1[/tex].You can check that by using the ε-δ definition of a limit.
And since f(x) is continuous in [x,x+Δx],[tex]f(x+\Theta_{1}\Delta x)[/tex] will go to f(x) ,and [tex]f(x+\Theta_{2}\Delta x)[/tex] will also go to ,f(x).
And by the squeezing theorem :
[tex]\frac{dv}{dx} = \pi.[f(x)]^2[/tex] and thus:
[tex]V=\pi\int^{b}_{a}[f(x)]^2dx[/tex]
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