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Topic: new way of finding volume and surface via the Cell theory 8th ed.:
TRUE CALCULUS; without the phony limit concept #119

Replies: 2   Last Post: Feb 24, 2014 9:32 PM

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 plutonium.archimedes@gmail.com Posts: 8,755 Registered: 3/31/08
new way of finding volume and surface via the Cell theory 8th ed.:
TRUE CALCULUS; without the phony limit concept #119

Posted: Feb 23, 2014 2:18 PM

new way of finding volume and surface via the Cell theory 8th ed.: TRUE CALCULUS; without the phony limit concept #119

Alright, I am ending the 8th edition of this textbook True Calculus and do not have time to revise or elaborate on this chapter of the 7th edition, so am going to include it without revision. Hopefully in 9th edition I have the time to fully revise.

Perhaps I found a new way of doing volume and surfaces in mathematics.

Perhaps I found an equivalent method of the triple integral.

Two functions in 2nd dimension allows us to obtain area of closed figures such as those found in Green, Stokes and Divergence theorems. But then to get to surfaces and volume as in Stokes or Divergence theorem, we use the z-axis to stack planes.

For a surface with area, the z axis stacks planes containing graphed function curves in the x and y axis and the stacking of those curves turns them into a surface with area. We can work backwards also. We start with a closed surface with area in 3rd dimension and we take a fancy saw that cuts the 3 dimensional space into planes. Our curved surface is still there only contained in a multitude of planes containing a line segments of the surface and when we stack those planes we retrieve our surface.

For a volume, the graphed functions in the x and y axis are closed loops and the stacking of planes of these closed loops builds a 3 dimensional object of volume. Here again we can work backwards and starting with the sphere in 3rd dimension we take the fancy saw that cuts 3 D into planes and the planes contain various sizes of circles and when stacking all those planes we retrieve the original sphere inside.

So when we do a triple integral, are we in fact doing a double integral of a closed surface and by stacking planes of these surfaces we end up with volume.

I am trying to think of how a z component refers to the proper plane that contains either a line-curve or a surface, or a point (tangent point). For the z axis would be the putting in sequental order the proper planes to build the figure.

So the z axis would be a function of sequential order of planes.

I may change the name in the future, but for now, I call it the Culmination Calculus Theorem. It is a play on the name of the Divergence Theorem, for it is from the Divergence theorem that I invent and discover this Culmination Theorem.

Let me get right into it, without side delay.

Mathematics has a huge problem of explaining why Calculus is a 2 dimensional phenomenon and not 3rd dimension. And that volume and 3rd dimension is where the Calculus of 2nd dimension comes to an end, or culmination.

The problem is obvious in my prior posts where I attempted to generalize the Cell Concept to 3rd dimension. In 2nd dimension, Calculus prospers and flourishes, however, when we enter 3rd dimension, calculus comes to an abrupt end and ends with volume.

CULMINATION THEOREM of Mathematics: The stacking of planes that contain area creates volume and is a unique process in mathematics.

Proof: What I am going to have to prove is that there is one and only one method of obtaining volume in mathematics, and that method is the stacking of planes which those planes have a closed figure -- cells, that has area and the volume is obtained by summing up the areas of all the stacked planes. It seems almost impossible to prove, due to the uniqueness espoused. But as I was waking up this morning and thinking about this, the uniqueness is easy to prove, because at infinity which is 1*10^603 of Floor-pi*10^603, the volume of the sphere and associated pseudosphere is exactly 1/2 the sphere. The pseudosphere trails in volume to the associated sphere, until it reaches Floor pi*10^603, then for an instant, the volume of the Pseudosphere crosses-over the numeric value of the volume of the sphere, due to the 3 digits of zero in a row in pi that allow the pseudosphere to gain more volume while the sphere volume continues to stagnate in adding more volume because the digits are 0 and thus 0 volume is added. The proof will show that as we make planes in 3rd dimension and those planes contain area as cross sections that as we disassemble the sphere planes to reform them into the pseudosphere planes. That the planes with area is the unique method that gives us exactly the 1/2 ratio of pseudosphere to sphere volume. QED

Now let me give a down to Earth example of what I am speaking of, so it is clear as to what is to be done.

We take the 10 Grid, for even High School students of little math can understand 10 Grid with its pretend that 10 is the infinity borderline.

We have the 10 Grid of 3rd dimension of its 1st quadrant only and we remove the other quadrants. So wee have only the positive x, y, and z of 1st quadrant.

We draw a half sphere in this 10 Grid, so that its center is (5, 5, 5). Its radius is 5 and its diameter is 10 and it is a hemisphere, for it is just 1/2 of a full sphere. The volume is equal to the pseudosphere.

Now we cannot draw the pseudosphere in the 1st quadrant because there are 2 arms, for we have a full pseudosphere but only 1/2 of a sphere.

Now in the 10 Grid there are 100 cells in this 1st quadrant of the x and y axis. They are 0 to .1 then .1 to .2, then .2 to .3, etc on up to 9.9 to 10. Each of those cells is in the x, y plane and to involve the z axis in 3rd dimension is that the z axis are planes of cross sections of the hemisphere.

Now look at a globe of earth and the lines of latitude and suppose we are in the 10 Grid, which means that we have a cutting block that would cut the globe into 100 cardboard thin slices. And we can take all 100 cardboard slices and reform the globe, or we can use those 100 slices to form a pseudosphere.

Now the slices are going to be different sized circles. There is one largest circle and there is a slice which is only a dot sized circle.

And here is where the Culmination theorem comes into play. We know at infinity, the hemisphere volume equals the pseudosphere volume. The only way to obtain that, the uniqueness, is that volume is these planar cross sections stacked. The hemisphere has 100 planes, and the pseudosphere has 200 planes. So that means we have to take some area out of the 100 planes of the hemisphere and create 100 new planes so that as we stack the 200 planes they exactly match the pseudosphere planes.

Now remember, in New Math there are no curves, but tiny straightline segments, so our hemisphere is more like a polygon, and as we take area out of the hemisphere planes to reform 200 planes, we are taking .1 by .1 squares and transferring them between one plane and a second plane.

No other method can do that, to arrive at the volume of the pseudosphere, and so the uniqueness.

Now let me warn the reader that the means of establishing the pseudosphere was 1/2 volume of the sphere was a parametric upon the sphere. The parametric is this process I described above of stacking planes of area to achieve a volume.

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Recently I re-opened the old newsgroup of 1990s and there one can read my recent posts without the hassle of mockers and hatemongers.