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Any one interesed in tasking on a formula for the volume of a sphere.

  1. May 8, 2013 #1
    An alternate one by the by. My approach will be working with a base ten logarithm function(s). I'm going to graph the sphere on a 3d graph. I'm going to give the value of the radius, one unit. I will first look at the ten elevations up and down from the index, establish the area at each elevation. Then see if I can identify any predictable relationship between each one. If it proves to be elusive, I'll work with 100 elevations, above and below. If this proves fruitful, I will look at the horizontal axis to see if I can predict the largest amount of perfect cubes that would fit within all the lines of the sphere, using whatever base I desire, as long as it is in relation with a base ten pre-ulate. Heh, coined a term, I think. In a sense this would comprise the regular cube that can occupy a regular sphere, except the ones that would stack up in descending number from the plane of the cube, to the plane of the sphere; yet I should understand the logarithmic function relation, which should provide me with a ratio of all that lies without the cube(s), and the additional smaller cubes, yet still lie within the sphere. This may have already been done, and I would be busting my B***s to no good purpose. I guess it would be referred to as 'cubing the sphere'; which, on the whole sound quite sexy. Any one wish me luck, on this endeavor?

    Jeffrey
     
    Last edited: May 8, 2013
  2. jcsd
  3. May 9, 2013 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    I think in the best case, you get some reasonable approximations for the volume of a sphere. Calculus can do this in an exact way - you can learn something interesting, but the result will not be relevant.
    Base 10 is nothing special. 10 just happens to be the number of fingers of humans. Systems with other bases were used, too, and modern mathematics in general is independent of any base.
     
  4. May 9, 2013 #3
    mvp;

    Thank you for your thoughts and discussion. I believe I understand your discussion on using the base ten, but for my level of ability with mathematics it would benefit me most, as the convenience of scale-ability could be accomplished with minimal brain overload on my part. ;) Once again thanks.
     
  5. May 9, 2013 #4
    You'd be better off thinking of the sphere (actually it's easiest to think of a hemisphere and double the answer) as a stack of ## n ## discs of decreasing radius. What is the volume of the bottom disc? What is the radius (and therefore what is the volume) of the next disc? What is the radius (and therefore what is the volume) of the disc whose bottom is a distance ## y ## from the bottom of the stack (note that by convention we would then say that each disc has thickness ## \delta y ##, and note that ## \delta y = \frac r n ##?

    Depending on how well your choice of discs approximates the hemisphere (clue: it works out best if the bottom disc is only half as thick as the others) you could be accurate to within 1% with only 5 discs. 16 discs gets you to 0.1%,

    Integral calculus gives us a way of working out what happens as ## n ## tends towards infinity, and so gives an exact formula for the volume of a sphere.
     
  6. May 10, 2013 #5
    MrAnchovy

    Thank you for the thoughts. Part of your discussion coincided with my approach. Even though I may not have expressed it well. In that regard I intend to just look at the area of each slice of elevation first, if I can identify the relation of each 'plane' circle first--each two-D slice--at the varied distances from the index; i.e. center of the sphere; then I will be ready to do the 'higher' analytical function(s). I plan to work with my 'native' intuition first and foremost. After which; irregardless of success or failure; I will look at the required 'disciplines' in order to learn, 'what I needed to know' in order to have solved the question smoothly and efficiently. Thank you both for your hints and ideas.

    God Bless spell check. Hey!

    Jeffrey
     
  7. May 10, 2013 #6
    If you can do a little math programing on a computer with a random number generator you could estimate the volume of a sphere.

    10 Start
    20
    30 count = 0, In = 0, out = 0
    40 Let x,y,z, be random numbers between -1 and 1
    50 If x^2 + y^2 + z ^2 is less then or equal to 1 then In = In + 1
    60 Otherwise out = out + 1
    70 count = count + 1
    80 If count = 1000 Goto 90 else Goto 40
    90 print Volume of sphere of radius 1/ volume cube of side 2 ≈ In/ 1000
     
  8. May 11, 2013 #7
    Dang you Archimedes; ya' took the thunder out of my approach. I'll have to think of some new and novel way now, I don't wish to be a geometrical plagiarizer now, do I now, indeed! Dang he's mind! Aw well, I suppose it was all known on the banks of the Ganges, long before they learned to walk there in Greece, anyhow. So, at least I'm coddled! With exclamation marks even. Take notice double null set.

    Thank you Spinner. As your suggestion seemed similar to the actual formula for a sphere. I decided to reference the actual formula, and discovered that an old dead Greek guy, had cubed the sphere long before I. Wonderful way to end a sentence! In rare form today, I do dare exclaim! Indeed!

    Wonderful things happen when you proofread, indeed. Wonderful! Now, only if there was a non-sense check. *sputters hot substance of a gaseous nature past 'es glottis.*

    'I didn't bother to read any further. ' It would seem;to this fool; that the subject phrase is; didn't bother to read..." That would make; 'didn't' the subject and noun, and 'bother' the modifier. So when the affected parties get done with there 'didn'ting, in their bothersome state, would they please see me after class.

    15 love

    I absolutely love matches made in heaven

    Bye now.
     
    Last edited: May 11, 2013
  9. May 16, 2013 #8

    Mark44

    Staff: Mentor

    This is unrelated to the purpose of the OP, but I couldn't resist. The subject of the sentence is the first person pronoun"I". The predicate is "didn't bother to read any further."
    Moved from Differential Geometry, as the question is unrelated to that branch of mathematics.
     
  10. May 16, 2013 #9
    Mark 44;
    Okay yes, you are exactly right, the subject is in fact I. Yet, I did not right that sentence, I was having a little casual fun with another. Worse yet; the quote was; 'I didn't bother to read further' which I added 'any' just for clarity. But you are correct, the 'I' of the of the construction is indeed the subject. In fact, it was the only point of the phrase. The OP could have left of with that and I would have been satisfied.

    Now, on the subject at hand, this Original post is something I will be getting at shortly. I'm hot on the concept outlined in Euclid's proposition 15 from book 4 of his 'Elements'. Seems I have some doubters, which unfortunately will be chagrined as I have already solved the 'proposition' [my proposition] less then 5 years hence. I developed a formula that demonstrated the validity of Pi, without using a square, or Pi, for all of that. I know all lot of you consider me to be a little dense, yet, think of what I think of you. Oh, by the by, my demonstration uses trigonometric functions. I'll trot it out for you all shortly. And then will post it to Beautiful Equations and expect it to stay posted.
     
  11. May 16, 2013 #10

    Mark44

    Staff: Mentor

    That's irrelevant. You commented on it, and that is what I corrected.
    As noted in another thread of yours, "5 years hence" does not mean "5 years ago," which is what I believe you are attempting to say.
    Back to mathematics...
    How can a formula demonstrate the "validity of pi?" ##\pi## is a number. If I were to write (note the spelling) 1 + 2 = ##\pi##, the use of ##\pi## in this equation would be invalid, but ##\pi## itself is neither valid or invalid.

    In your OP, you talked about determining the volume of a sphere by making horizontal slices. This sort of thing is done in calculus classes when definite integrals are the topic being discussed, so I'm not sure what the point is. I'm hopeful that you will discover the volume to be 4/3 ##\pi## (since you are using a radius of 1 unit).
    Speaking for myself, that's not something I give a great deal of thought to.
     
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