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I know that in a classic Cartesian coordinate system $xOy$, if I have a function $y = f(x)$ and I want to find the volume of the a solid of revolution around x-axis I can compute: $$V = \pi \int_{a}^{b} f(x)^2 dx$$ But, why?

Can, please, someone provide me a not-formal demonstration?Just an idea?

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Take the solid. Chop it up into "disks" that are perpendicular to the $x$-axis. The volume of each disc is $\approx\pi[f(x)]^2\cdot \Delta x$. Add up the volumes of the discs. The integral is what results as the number of discs tends to infinity. – David Mitra Apr 26 '12 at 19:16
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The key point (which I missed somehow the first time I took calculus) is that for any function $g(x)$, we have by definition $$ \int_a^b g(x) dx = \lim_{\text{as the partition gets finer and finer}}\sum_{i=1}^\text{size of partition} g(x_i)\Delta(x_i) $$ where by definition a partition of $[a, b]$ is a choice of $n+1$ numbers in a row on the interval with $a = x_0$ and $b = x_n$ and $\Delta(x_i)$ is short-hand for $x_i - x_{i-1}$. There are technicalities here - define a "finer" partition carefully - describe for which class of functions the limit exists, etc. None of these technicalities arise when computing volumes of solids of revolution.

Now in your case, you want to revolve this region about the $x$ axis. Pick a partition of your interval (pick an evenly spaced one with $n+1$ numbers, so that $\Delta (x_i)$ will be constant - I'll just call it $\Delta(x)$; it's just some small number, namely $\frac{b -a}{n}$.) Slice your region into thin vertical strips along the partition. We will pretend the function $f(x)$ is constant on each vertical strip - let's say it's $f(x_i)$ between $x_{i-1}$ and $x_i$; this is just an approximation, but it's a good one if your partition is thin.

Now when you revolve your strip, under our approximating assumption that $f$ is constant on your strip, you're going to get a cylinder (turned on its side). The radius of your cylinder will depend on which strip you took - it will be exactly $f(x_i)$, but the height (since the cylinder is on its side, I mean the very thin "thickness"), will be $\Delta x$. Therefore the volume of the cylinder will be $$ \pi f(x_i)^2 \Delta x. $$ Therefore the entire volume is approximately $$ V \approx \sum_{i = 1}^n \pi f(x_i)^2 \Delta x $$ The approximation obviously gets better and better as the partition gets finer and finer, so to get the actual volume, you can take a limit over all the partitions. But then by definition, this limit is the integral that you want!

Note that this is a miracle. Each individual approximation via a fine partition would be nightmarishly tedious to compute, but if you're lucky and $f^2$ has an antiderivative, the fundamental theorem of calculus magically finds the exact volume for you. I missed this point as a calculus student.

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really thank you and also I congratulate you becouse your explanation was very clear. – Aslan986 Apr 26 '12 at 19:55

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