Alright my 231st book of science is a brief history of Calculus-- true calculus. And in that book I want to offer a new method of integration based on the idea that stacking identical circles over a distance gives the volume of a cylinder. But stacking dissimilar circles like that of latitude lines in a sphere should also give the volume of the sphere as 4/3*pi*r^3. I am making some progress on this feat but not enough.

Notice the sphere surface area is 4*pi*r^2, meaning that 4 great circles of the sphere is enough to cover the surface of that sphere.

So here, I am looking at the Amount of surface area to compose volume.

The amount of surface area to compose a cube is 6*side^2 and the amount to compose a sphere is 4*pi*r^2.

What if we rewrite volume and surface area of sphere using just diameter, not radius. Then we have surface area of sphere as pi*diameter^2.

There is no getting around 6*side^2 in cube, but for sphere we have pi*diameter^2.

AP

Archimedes Plutonium
5:11 PM (2 minutes ago)
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The first calculus on Earth was the notice that if you multiply a length of similar circles over a distance you end up with the volume of a cylinder. This was the integral of Calculus. And we do the same method on boxes to get volume.

But until now, we could never do that method on spheres or cones whose circles have a gradient of area, not all the same as in a cylinder.

So what AP is doing is solving that method for volume of cone and volume of sphere, both of which have a gradient of circle areas stacked.

The key to solving this problem is likely to be how much surface area a figure can have to hold volume.

AP

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Archimedes Plutonium<plutonium.archimedes@gmail.com>
7:30 PM (27 minutes ago)
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Alright, all of these geometry figures are related to one another. The Rectangular box that houses inside itself the Cylinder, which in turn houses inside itself both the Sphere and the Cone. Where 2/3 of cylinder volume equals sphere volume and cone volume is 1/3 cylinder volume and semisphere volume equals 1 cone volume.

The picture for surface area is a bit different and is what I will use to form a gradient of circles over a length to solve cone volume and sphere volume.

The right-circular cone surface area is S.A.= pi*r^2 + pi*r*(sqrt(height^2 + r^2))

In contrast, the cylinder surface area containing the cone inside it is S.A..= 2*pi*r^2 + 2*pi*r*height

Let me change both of those surface areas to diameters which can probably make more clear the task at hand-- volume from stacking different sized circles.

cone S.A. = (1/4)pi*diameter^2 + (1/2)pi*diameter*(length of slant line of side wall) = (1/4)pi*diameter^2 + (circumference of base circle)(length of slant line of side wall)

cylinder S.A. = (1/2)pi*diameter^2 + (circumference of base circle)(height of cylinder)

Now it would appear that the slant height in the formula for cone surface area is a mistake comparing to the associated cylinder side surface area.

Perhaps this is a grave mistake of Old Math. Another time about 5 or 10 years ago I wrestled with the torus problem formula and was finally able to clarify Old Math's averaging of one of the radii. But there is no Averaging here with Cone, and so I suspect Old Math formula of cone surface area is found to be in error. But let me do further checking.

AP
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Archimedes Plutonium<plutonium.archimedes@gmail.com>
7:49 PM (7 minutes ago)
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Yes, huge huge error in Old Math-- their formula of surface area of right circular cone pi*r^2 + pi*r*(sqrt(height^2 + r^2)) for if you take a plastic cylinder as what I have and a plastic associated cone and set them side by side, it is obvious the cylinder side wall is far larger in area than the cone side wall.

When I had a fight of this kind many years back over the torus formulas, what saved the Old Math formula from ruin was a "Average Radius. But in the case of the cone, there is no averaging involved and so the Old Math Formula is in ERROR.

AP
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Archimedes Plutonium<plutonium.archimedes@gmail.com>
7:55 PM (1 minute ago)
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And this error of Old Math is easily seen by a wand stick, as it goes around the cylinder, it accurately measures the amount of area of the rectangle as cylinder side. But the wand stick as it goes around the cone has severe problems with the apex and the base.

AP

Sorry, AP apologizes for the mistake on his part, for in the cone formula it is only a 1/2 circumference, a sort of Averaging that saved Old Math formula of Cone surface area, whereas in Cylinder formula it is a full circuit around the base circle. For cone it is only a 1/2 circuit.

So the Old Math formula is correct, AP was wrong, but always alert, for I love to show Old Math for the lack of marbles they have.

AP

Sorry, not in deep error but error in explanation. They have the correct formula of surface area but in error of its true explanation.

And this true explanation will by my 232nd book of science-- True Explanation of the lateral area of Right Circular Cone.

The Internet has the explanation all messed up showing some pictures with the cone cut away as being /\circle and some showing it as \/circle.

The Quora does the correct algebra but fails on the cutaway picture.

The Byjus does a correct picture but fails on the algebra.

So what AP offers in this correction is that AP notes this is the first encounter in mathematics where a PERCENTAGE of one area by comparing to another area takes place. I wrote one of my books saying the Percentage method of extracting volume or area is a major method of geometry.

But here on the Cone lateral surface area, to obtain that formula we place the cone insided a circumscribing cylinder and take the cylinder surface area as 2pi*r*height and compare in ratio to the cone's singular circle base of 2pi*r*slant-height.

Now, I need to check to see if any Old Math Calculus can derive the formula of pi*r*slant-height for lateral surface area.

I am guessing no Old Math calculus can sneak in the idea of a Percentage Comparison of the cone with connected associated cylinder in order to fetch the true formula.

In this regard, AP has found a mistake in OId Math of its cone-- the inability to recognize the true proper explanation-- a Percentage comparison. And as for the drawings the 1/1.4142... = 0.707 or about 71%.

AP