Here is another approach to finding the volume :(adsbygoogle = window.adsbygoogle || []).push({});

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]........................................................................................1

Now 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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# Volume, washer method another approach

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