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Today I learnt about how the disc method can be used to find the volume of rotating solid.

It is $$\pi \int_{a}^{b} (f(x)) ^ 2 dx$$

The proof for this uses the integral to calculate circle areas and the integral sums them up.

But I can't understand how this works.

It assumes a the area to a be a circle but it isn't a circle. Even if we go infinitesimally small, I don't think we can approximate like that. After all even really small errors in approximation can add up to big errors.

Can anyone please help me explain the proof for this?

In general, how exactly do we know when to approximate something and when to not?

Sorry if this is a stupid question :P

Thank you

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MangoPizza
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  • Are you happy with using rectangles/Riemann sums to approximate the area of a plane region enclosed by a graph? – Andrew D. Hwang Apr 02 '22 at 18:47
  • The Wikipedia page doesn't seem to actually show any of the disks that give the "disk method" its name. The figure in the question is just a planar figure with no volume at all; the volume comes from rotating a region around the $x$ axis. You don't even quite have a region in that figure, let alone any visualization of the rotation, so you would really have to have a good imagination (or already know how the disk method works) in order to see a "disk method" in this figure. – David K Apr 02 '22 at 21:54
  • There are authors who actually show the disks of the disk method (and not just their circular faces), for example, https://www.sfu.ca/math-coursenotes/Math%20158%20Course%20Notes/sec_volume.html – David K Apr 02 '22 at 22:02
  • @DavidK My question is not about the rotation. The disk method assumes the volume to be very small circles, but even when we go really small, the volume of $dx$ enclosed graph, when rotated, won't be a cylinder. Thanks – MangoPizza Apr 03 '22 at 04:15
  • See my figure, if we rotate the $dx$ enclosed graph, the figure formed would not be a cylinder. (But in the integration we assume it to be a cylinder) I think that in reality a conical frustum would be formed. – MangoPizza Apr 03 '22 at 04:29
  • You're right, a conical frustum is a better approximation. For the purpose of surface area, in fact, the frustum is a good approximation but the cylindrical disk is not. So this is a very good question. There are possible answers here and here; do they help? Or would it be better to have an answer devoted just to volume without the distraction of surface area? – David K Apr 03 '22 at 23:23
  • @DavidK A conical frustum is a better approximation, but it's not the accurate approximation. After all the edge in a conical frustum is straight, but this is not guaranteed here. That's why I was interested in a proof for disc method that does not approximate. After all, small in-accuracies can add up to big ones. (I hope you're getting my point, this is difficult to explain) Thanks – MangoPizza Apr 04 '22 at 12:56
  • I also saw on another Wikipedia page this proof $V = \iiint_D dV = \int_a^b \int_{g(z)}^{f(z)} \int_0^{2\pi} r,d\theta,dr,dz = 2\pi \int_a^b\int_{g(z)}^{f(z)} r,dr,dz = 2\pi \int_a^b \frac{1}{2}r^2\Vert^{g(z)}_{f(z)} ,dz = \pi \int_a^b f(z)^2 - g(z)^2 dz$ I don't understand what this means, but I think this proof doesn't approximate to cylinders, frustum. – MangoPizza Apr 04 '22 at 12:58
  • Link: https://en.wikipedia.org/wiki/Solid_of_revolution#Disc_method – MangoPizza Apr 04 '22 at 12:59
  • @DavidK Maybe, I think that indeed a conical frustum would be formed. After all, when computing derivatives, we consider the slope of the graph at some point, so we assume the graph to be straight line infinitelsmy small. In that case, volume = $\int_{a}^{b} 1/3 \pi (f(x + dx)^2x + f(x + dx)^2 dx - f(x)^2 x)$ How to simplify this further? Thanks. – MangoPizza Apr 05 '22 at 13:29

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