Volume, washer method another approach

[evagelos]

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.........

$$\pi.[f(x_{1})]^2\leq\frac{\Delta v}{\Delta x}\leq\pi.[f(x_{2})]^2$$................1Now 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:
$$V=\pi\int^{b}_{a}[f(x)]^2dx$$

 

[jvicens]

Thank you for sharing this approach! It seems like a clever way to use the minimum and maximum values of f(x) to approximate the volume. However, as a scientist, I would suggest considering the limitations of this method. Firstly, the accuracy of the approximation depends on the choice of x1 and x2. If they are not chosen carefully, the error in the approximation could be significant. Secondly, this method assumes that f(x) is continuous, which may not always be the case in real-world scenarios. It would be interesting to explore how this method compares to other methods for finding volume, such as using Riemann sums or numerical integration. Overall, this is a great contribution to the discussion on finding volume and I appreciate your use of mathematical concepts like the squeezing theorem. Keep up the good work!