# Volume, washer method another approach

Here is another approach to finding the volume :

Let $$x_{1}$$ be a point in [x.x+Δx] such that :

f($$x_{1}$$) = minimum value of f(x) in [x,x+Δx]

Let $$x_{2}$$ be a point in [x,x+Δx] such that :

f($$x_{2}$$) = maximum value of f(x) in [x,x+Δx]

Then,

$$\pi.[f(x_{1})]^2.\Delta x\leq\Delta v\leq\pi.[f(x_{2})]^2.\Delta x$$

..........................OR..........................................

$$\pi.[f(x_{1})]^2\leq\frac{\Delta v}{\Delta x}\leq\pi.[f(x_{2})]^2$$........................................................................................1

Now let :

$$\Theta_{1}=\frac{x_{1}-x}{\Delta x}$$ ,and

$$\Theta_{2}=\frac{x_{2}-x}{\Delta x}$$

..................and (1) becomes:

$$\pi.[f(x+\Theta_{1}\Delta x)]^2\leq\frac{\Delta v}{\Delta x}\leq\pi.[f(x+\Theta_{2}\Delta x)]^2$$.

And as Δx goes to zero,

$$x+\Theta_{1}\Delta x$$ goes to ,x

$$x+\Theta_{2}\Delta x$$ goes to ,x, because $$0\leq\Theta_{1}\leq 1$$ and

$$0\leq\Theta_{2}\leq 1$$.You can check that by using the ε-δ definition of a limit.

And since f(x) is continuous in [x,x+Δx],

$$f(x+\Theta_{1}\Delta x)$$ will go to f(x) ,and

$$f(x+\Theta_{2}\Delta x)$$ will also go to ,f(x).

And by the squeezing theorem :

$$\frac{dv}{dx} = \pi.[f(x)]^2$$ and thus:

$$V=\pi\int^{b}_{a}[f(x)]^2dx$$

Last edited: