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let $w$ be the width of a rectangle and $l$ be the length. From the definition of area it sounds logical to add up $w$ lengths of a rectangle to find how much two dimensional space it is occupying. When we carefully see what we did we take a one dimensional quantity and and added it some $w$ amount of times to get the area of this surface. Can't we do the same with three dimensional bodies. Doesn't it sound intuitive to slice up a revolution(a 3d object) into many 2d circles and find the sum of their perimeter to find the surface area of this 3d object.

*I am sorry I dont know how to properly put my idea in chronological order or use Latex to properly write a proof so I attached an image of my question. I would be happy if any one edits my question and writes it in proper form.

My question

The thing is I am wrong. To show that I am wrong we can do a proof by contradiction. let $f(x)=x^2$. By the already established method the surface area from $0$ to $a$ would be different from what my method would get which is infinity. I will put what I said in word in the next picture.

the illustration

So my question is Why am I wrong?, and why don't we use this intuitive idea?

EHM
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  • @user729424 thanks for the edit – EHM May 02 '20 at 12:56
  • I think this Wolfram page about the Diagonal Paradox will shed some light on the problem. It concerns arc length and not surface area, but the idea is similar. – TonyK May 02 '20 at 13:04
  • In the illustration, the sum $\sum_{k=0}^{n} 2\pi \cdot f(\frac{ka}{n})$ is adding the circumferences of circles. So, here what you're calculating is more like how much wire would wrap around the surface (?). If you want to find the area, you should be adding up areas of small cylinders instead, each with a height $1/n$ (here, the "height" is measured horizontally). The sum of these areas would then be $\sum_{k=0}^{n} 2\pi \cdot f(\frac{ka}{n}) \cdot (1/n)$. – ZxJx May 02 '20 at 13:20
  • @ZxJx But ya that is exactly my question is. (you put it in a nice analogy) wouldn't calculating the total wire i need to wrap up the surface of solid like the one I mentioned equal to the total surface area. Basically if I unwrap the three dimensional surface into a two dimensional figure, then cut it into a fine thin threads wouldn't I get the amount of wire I need to wrap it up. – EHM May 02 '20 at 13:36
  • My point perhaps is better put this way: when you want to calculate surface area, make sure you're splitting the surface up into smaller surfaces and adding those $areas$. You want to add smaller $areas$ to get the total $area$. The argument in the summation in The Illustration is measuring a $1-dimensional$ amount ($length$ only), not a $2-dimensional$ measure ($length$ and $width$). – ZxJx May 02 '20 at 14:01
  • I get that we have to take smaller areas and add them to get the total surface area. I don't have a problem with the conventional method. But i am asking what the problem is with taking a smaller dimensioned figure and adding it up to get the area of a biger dimension obkect. Like we take a single dimension (line segment) and add it up to get the area of a 2 dimensional figure like a rectangle. – EHM May 02 '20 at 15:13
  • See this related answer https://math.stackexchange.com/a/1692595/72031 – Paramanand Singh May 02 '20 at 15:54

1 Answers1

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You can find the area of a solid of revolution by slicing it into circle and adding up the circumferences. But you have to be more judicious about how you slice.

The idea in the comments about wrapping wire rings around the surface is a good one. The figure below shows two surfaces of revolution: a cylinder, and a cone inside the cylinder.

enter image description here

All the wire rings are made from wire of the same thickness. But in the same distance along the axis of revolution in which we can fit five wire rings around the cylinder, we can fit seven wire rings around the cone.

Conceptually, it's very simple. Take the curve from which you generate the surface of revolution. Mark off equal distances along that curve and put a circle at each mark. Now the sum of the circumferences times the distance between circles along the curve is a reasonable approximation of the surface area, and becomes an accurate measure of the area once you turn it into a Riemann sum and take the limit as the distance goes to zero.

What you must not do is take circles whose centers are spaced at equal intervals along the axis of rotation and treat their circumferences as wires wrapped around the surface. The circles will be too far apart in the tapered sections of the surface. Consider what would happen if we replaced the seven rings in the figure about with five rings around the same cone; there would be gaps between the rings, and we would not be covering the entire surface.

In practice, marking off equal distances along a curve is not always easy. So instead we usually put the centers of the circles at equal distances along the axis of revolution, but multiply each circle's circumference by the distance between the circles along the curve, which is a larger distance in the tapering sections of the surface than in the cylindrical sections.

David K
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  • "What you must not do is take...." why not? As long as i will be taking the limit as n approaches to infinity. Wouldn't the gaps be filled as n approaches to infinity – EHM May 02 '20 at 15:00
  • And even if this method gives a less surface area as you have said why am I getting infinity. Shouldn't I be getting something less than what the conventional method give me. – EHM May 02 '20 at 15:05
  • If your wires only cover $70%$ of the surface when their thickness is $0.1$, then they will still cover $70%$ of the surface when their thickness is $0.01$ or $0.0001$ or $0.0000001.$ The gaps between the rings get smaller, but the number of gaps gets larger. – David K May 02 '20 at 15:17
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    As for how you get infinity, I misunderstood your method at first due to the almost illegible pictures. I thought you were making the usual mistake of trying to adapt the disk method to area measurement without accounting for the $\sqrt{1+(f'(x))^2}$ factor. But you were literally just adding the lengths. This is never a correct way to measure area under any circumstances. Area is always length times width, so your "wire" should have a non-zero width so that it actually covers the area of the surface. As the number of circles gets larger, the width of the "wire" must get smaller. – David K May 02 '20 at 15:46