Volume of Sphere & Calculus: From a layman's point of view S V Ramu (2000-11-20 -Version 2.0-)
Have you ever wondered about the volume of Sphere? Without
Calculus?! Sphere has been a solid of awe, from the time
immemorial. It is the perfect
We will try to find its Volume in a pristine way, in an
attempt to recapture the original impetus that drove
mathematicians and laymen alike, to discover the nexus between
It was Xeno, who first expressed the enigma of
This story is almost clichéd, yet the inspiration behind it is as fresh as ever. In fact the details of this story can be varied and yet the paradox will remain. For example, the handicap distance can be anything other than zero; it doesn't matter! The Tortoise could have been a snail! Let us analyze and solve this riddle (an obvious fallacy isn't it?). To that end let us assume some values to aid our calculations (it is easy to prove that our assumptions are inconsequential). Lets say, that the tortoise is half as fast as Achilles (of course preposterous, but this will greatly simplify our calculations). So, it is one meter ahead at the beginning, and it moves half a meter while Achilles cover that one meter, and then quarter of a meter and so on. The total distance covered by the tortoise will be,
Now the puzzle is, if the sum of this series is infinity, it
means that Achilles can never win, because then, Achilles would
be forever catching up with the tortoise! But, of course this is not true
in reality! So, what is wrong? See closely! Does the sum of this
series equals infinity? How to prove this, without resorting to
some hi-fi math? But, how can the sum of infinite things (however
small), can be anything but infinity?! after the
infinite steps, the total size in the pocket will be
exactly equal to One! Thus,
Hence the
The resolution of this paradox was the first step in
comprehending the nature of infinity. From here on, the
possibility of
Let us push this Let us look at the sum of first 1 + 2 + 3 + 4 + 5 + ... + n = ? What is the sum of this series? One thing is clear; the value
of the sum should only depend on 1 + 2 + 3 + 4 + 5 + ... + n = n(n+1)/2 Many of you should have seen this. But how is this true? How
can the human mind conjure such fantastic short-form for a
lengthy series? Remember this 1 + 2 + 3 + ... + 98 + 99 + 100 = 100(100+1)/2 = 5050 There is an interesting anecdote with respect to this problem. It is said, that Galois was given this problem when he was 10 years old. Apparently, the teacher of his class was worried that the kids were too restless and bored. So, she asked them to sum all the numbers up to 100. As they had learnt to add and subtract only recently, she expected them to take long time to finish this task. Not true for Galois! He noticed an interesting pattern. Consider the following, 1 + 2 + 3 + ... + 98 + 99 + 100 = S (say!) also, 100 + 99 + 98 + ... + 3 + 2 + 1 = S (of course!) Now, what is so interesting in this? Let us see... 1 + 100 = 101 This is the beauty! Adding 1 to 100 in either way should result in the same sum, right? So, if summing 1 to 100 equals S, summing 100 to 1 (in reverse) should also be S. Hence, summing both together should be equal 2S, right? That is... (1 + 100) + (2 + 99) + (3 + 98) + ... + (98 + 3) + (99 + 2) + (100 + 1) = 2S Thus, 101 + 101 + 101 + ... (100 times this way) = 101 x 100 times = 10100 = 2S Simplifying we get, S = 10100 / 2 = 5050 Did you see it! How can we ever find reasons for the
astounding beauty of a genius's mind! Galois made the problem look so
simple, that his teacher was spellbound. What is more, this method
was found to be holding true for 1 + 2 + 3 + ... + (n-2) + (n-1) + n = S (say) Now adding both, (1 + n) + (1 + n) + (1 + n) + ... (n times this way) = n(1 + n) = 2S Simplifying, S = n(n + 1) / 2 Just see and praise! This derivation needs no commentary. Its
elegance speaks for itself! Now, a seeking mind will not stop at
one glory. It will push further. A true mathematician's mind
enjoys Here we go on to, 1² + 2² + 3² + ... + n² = S (say) Again I would like to tease you with the answer, before going
to its roots. I find this an effective way to learn something.
The awe should precede all our search for knowledge. Only
1² + 2² + 3² + ... + n² = S = n(n+1)(2n+1) / 6 Complex eh! Just wait. We want to find the sum of all squares from 1 to n. Some times a method that was successful hitherto may not be applicable to a problem in hand. This can be so, not only in math, but also in other aspects of life. We must understand that this is neither preposterous nor ugly. With what assurance can we say that our efforts are absolute? How can we assume that the first luck we stumbled upon as the ultimate truth? These ramblings are not some philosophical sophistry. It is
said that Newton wrote down some In this vein, we shall But first, the sum of natural numbers again, to arrive at some broader technique! (Sorry! We have to see the same thing in different angle to get a firmer grasp of it). But, don't worry; I'll try to make it as intuitive as possible. Be warned, it is going to be wordy! (Or, should I keep it plain?) We know, (n + 1)² = (n + 1)(n + 1) = n² + 2n +1 Now let us do a mental leap. You must remember that the
awesome tricks that mathematicians do are not the product of one
day. Some of the seemingly simple methods that we see today have
decades or even centuries of history. It is said that the
So, if you ever wonder how some ingenious way was found by a great mind, remember it did take its toll of time and history. Just try your best to relish and enjoy it, and you are on your way to some personal landmark! All this said, follow on... 1² = 0² + 2.0 +1 Summing both sides of all the above equations, 1² + 2² + 3² + ... + (n+1)² = [0² + 1² + 2² + 3² + ... + n²] + 2.[0 + 1 + 2 + 3 + ... + n] + (n + 1).1 If you notice, all the squares on both sides cancel each other except (n+1)². So, (n+1)² = 2.[0 + 1 + 2 + 3 + ... + n] + (n +
1).1 where 1 + 2 + 3 + ... + n = S (the same thing we are finding again!) Thus, 2.S = (n + 1)² - (n + 1) After some thought, obvious isn't it? But see its elegance too! The key feature of this method, against the Galois's masterstroke is its generality. You can extend this method for squares, cubes and more. Like, We know, (n + 1)³ = (n + 1)(n + 1)(n + 1) = (n² + 2n + 1)(n + 1) = n³ + 3n² + 3n + 1 Now go on as before. Sorry, against my will, I'm not going to go through this fully again. But if you are not convinced that,
Then, please work your way through this fully. In fact, when I
did that in my early days, I found this very complex (though
solvable) for squares and cubes (not to mention higher powers).
But, as providence usually gives complexity for growth, I did
find an elegant method for 4 Now that we are armed with all these results, we shall boldly
proceed to solve the
You all should have heard the story of What does all these mean to our understanding of
We saw that,
This is
As, Limit n tends to infinity, of
Clear? Ok, now the problem is, how do we say what the sum of
an infinite series would be when its terms steadily decreases?
Here the term First off, remember, we haven't said that all infinite series
is
So, we have to be cautious. How? Let us clear some basic
stuff. Which is bigger? Limit n tends to infinity, of
Aha! That is something! Now we turn the question on its head. What happens if the denominator becomes smaller and smaller?
So, the value of the fraction increases, as the denominator decrease. How big can the fraction be? As big as we want! Yes, as the denominator goes to zero the value of the fraction becomes larger and larger, i.e. It becomes infinity. Thus, Limit n tends to zero, of
Great! We are getting somewhere. To simplify things,
And,
Wow! Let us see what we can do with these results. For
example, let us consider some ugly expression like (n + 1)(2n +
3) / 5n² as Limit n tends to infinity, of (n + 1)(2n + 3) /
5n² Now, making the Thus, Limit n tends to infinity, of (n + 1)(2n +
3) / 5n² = ²/ Do you see the power! We shall need these analysis and
techniques in finding the
Now you are going to witness a powerful technique, which trace back to a time, long before that of Newton's. The idea is to split a continuous thing into infinite number of discrete things. This is a very natural idea. Imagine, you want to find the area of an irregular figure. How will you do it? One nice idea is to divide the irregular figure into a grid of small squares and then add them up. The smaller the squares, higher your accuracy will be. If the area of squares tends to zero, then the number of squares tend to
infinity, and hence the area is exact!
Here, divide one half of the sphere into R² = r² + (k .
R/n)² Now, what is the Volume of the Whole Sphere? = 2 times the volume of each hemisphere You should understand that each disk is a cylinder, and, Volume of a cylinder = Area of its base
Thus the = 2.{π [R² – (1.R/n)²].(R/n) + π [R² – (2.R/n)²].(R/n) + .... } Pulling out the π and (R/n), = 2 π (R/n).{ [R² – (1.R/n)²] + [R² – (2.R/n)²] + .... + [R² – (n.R/n)²] } Expanding and Pulling out the common R², = 2 π (R³/n).{ [1 – (1/n)²] + [1 – (2/n)²] + .... + [1 – (n/n)²] } Summing the common terms, = 2 π (R³/n).{ n.1 – [(1/n)² + (2/n)² + .... + (n/n)²] } Pulling out the common 1/n² we get, = 2 π (R³/n).{ n – (1/n)² [1² + 2² + .... + n²] } Now, using the
= 2 π (R³/n).{ n – (1/n)² [ n(n + 1)(2n + 1)/6] } Pulling out the common factors and Simplifying, = 2 π (R³/n).n{ 1 –
(1/n)² [(n + 1)(2n + 1)/6] } Well! This is the Thus when the limit of n tend to infinity, in the above final
equation, the 2 V = 2πR³.{1 – 1/3} =
(4/3)πR³ At last, we have found the Volume of Sphere! Note it, we haven't used any Calculus! Now, the only thing that remains, is to show the Newton's effort in simplifying this whole process through Calculus.
Enter Newton! And this whole process was simplified through Integral Calculus. The beauty of the above algebraic method of finding the Volume of Sphere is, it doesn't use any unsaid concept. In fact you can, if you want, altogether abolish calculus, and use this technique of summing an infinite series over a limit for all your
needs. But that would be rather painfully tedious, since there
are some simple patterns in the concept of limit itself, that we
can use, to simplify. Remember, the jargons and techniques of
calculus are just, short forms for summing an infinite series
over suitable limit. No more, and no less.
Now, let us finish off the holy grail of any higher secondary
student, that of finding the relation between the ordinary
algebra, and calculus. To me, this facet took lot of time to be
intelligible. In retrospect, the trouble is in coming to terms
with the obvious. The answer was really right before my eyes, yet
I didn't believe, as it looked too simple to be true. Yes,
calculus is really a short form for Very simply put, integral f(x)dx, between x=a and x=b (b>a)
can be
If you notice, the same trick that we applied to the above case of sphere holds good for all continuous functions. For the area under the curve case, assume that the x-axis is split into n equal sections, between x=a and a=b. So, each section (say h) is equal to (b-a)/n. h = Now visualize rectangles with breadth as h, and length as f(h), f(2h), ... and f(nh). In the spirit of limits, can we not then say that, the sum of the areas of all these rectangles, as the area under the curve between x=a and x=b ?! Of course, if the number of sections (n), thus the rectangles, is infinitely large. So,
where
h=
^{(b-a)}/_{n}
_{a}∫^{b} f(x)dx = ^{lim}
_{n
→ ∞} h. , _{k=1}∑^{n}
f(k.h)where h=
^{(b-a)}/_{n}, since h is
a constant with respect to rWell, this is powerful! Do you realize that? If you don't I
will not blame you. (Beware, there is a mistake here, that we'll
clarify in the next section. It is ok for now!) For example, I
was almost blind, or definitely hazy, about its value for many
many years. What the above equation defines, is a technique to
interpret the
And, _{0}∫^{R} f(x)dx = h.
_{k=1}∑^{n} f(k.h),The question is, what should be the f(x) in sphere's case? If we find that, we can, not only relate integral to volume of sphere, but also cast our derivation in the language of calculus. Remember, until now, we are only using calculus as notation shorthand, and nothing more, hence the derivation will not be any simpler. Once we frame this connection beyond doubt, we'll see some properties of integral, which allows many problems like that of deriving volume etc. to be dramatically simple. For now, what is f(x) here? So we know, Here, since h acts as the height of the cylinder, f(k.h) is
nothing but the area of the circle of k Now, Area of the Circle = π.radius² Isn't this wonderful! The integral form of the volume of sphere is,
Well! I suppose the simplicity speaks volumes for itself. See the amazing brevity of saying the same thing, that we previously said in so many words, in so little. Of course, expressing something in a cute integral form doesn't solve it. All the same, we do know that if we can readily know the integral of all functions, and can cast a given problem into an integral of an appropriate function, then the solution is not only immediate, but also most elegant. By the way, in case you want to see the steps from here on, Volume of Sphere = 2
The last section's goal was to relate the modern Integration
techniques with the bygone Indefinite
Integral. If Definite Integral (i.e. between limits a and b)
means the area under the curve between the x=a and x=b, then what
does Indefinite Integral (with no lower or upper limits
specified) mean? This is an important question. As you will see
the answer is obvious (in hindsight). The deeper doubt in this
question is, if we are to assume only the definition that
Integration is just a handy abbreviation for the summation of
some suitable convergent series, and not use any of the
modern results of integrals (which use differentiation and other
tools, hitherto un-discussed here), then how are we to define a
limit less Integration in terms of summation?!
In the last section, we said,
where
h=
^{(b-a)}/_{n}
where h=
^{(b-a)}/_{n}
Now, coming back from this little digression let me give out the answer before going into the motivation behind that, as is usual. ∫ f(x)dx = Ok, what is the point in making such a definition for definite integral? If, ∫ f(x)dx = To understand that, see the figure in this section. You can notice that, (The area between x=a and x=b under f(x)) = (The area between x=0 and x=b) - (The area between x=0 and x=a) Of course assuming that a<b, we can reduce the problem of definite integral to one of subtracting areas. A powerful result, always reminding us of the geometric origin of the definition. In calculus, as in other fields, it is very important that we constantly be aware of the definitions and other assumptions, and not be carried away by the complex looking expressions, steps and symbols.This reminds me of an anecdote of Feynman (a noble laureate in physics). It seems that he was giving a lecture to a distinguished audience which had Einstein, Bohr and such like. Feynman was uncomfortable with the physical interpretation of a step in his derivation. He was already legendary for his superior usage of mathematics for physics. Moreover his lecture was planned for a couple of hours. So, he thought, that this miniscule step could be overlooked for now. Thus, the lecture went on and finished well. The question time started. It seems that the very first question from Einstein was about the doubtfulness of the very same miniscule step! Feynman was astonished. He asked Einstein (already a revered octogenarian) how he could single that very problem out of a couple of hours of intense deep-math lecture? The reply was, that mathematics is only an expression of the physicist, whose primary objective is to model the given physical problem. (I'm truly not worried about the authenticity of this anecdote, but am willing to subscribe to the fact that, as long as we are fully aware of the motivation and significance of a symbol, a definition, an axiom, or a concept, nothing else matters. A derivation or proof is only an expression of our inner conviction that something is true. Long before a proof or a theory is spelt out, the mind has already seen it. Mind doesn't use symbols and expressions alone to come to a conclusion. Understanding is much more than knowledge.) Another question that haunt us is, what is the significance of dx? The answer is: just to hint us what is the variable with respect to which we are integrating. Yes, in the modern method of finding the value of an Integral, dx does not play any active role other than tagging the variable of integration. But for formulating a real-life problem, from convergent series notation to integral, this dx is the buoy, which hints us about the whereabouts of the breadth of the rectangles of summation (lengths being the function f(x)).
Up till this alone is enough to satisfy a high schooler about the real-time relevance of calculus. Since they already would have gone through the mill of finding integrals, without knowing how it relates to the problem at hand. Still, one more thing can be of good use. If you remember, we used summation of series, and limits, to calculate the volume, and hence the integral. But, if we have to do that always, that looks bit unwieldy. It is! I heard that it was indeed done this way sometime back in history, but latter an amazing inverse relationship was found with Differentiation (another facet of calculus, ie. limits).
Differentiation happens to be a rather simple in computation, and
hence asserting that Integration is inverse of
Differentiation, reduced lot of efforts. Almost always you
can just manipulate a differentiation in reverse and find the
corresponding Integral value. If you need to know how a given
integral is derived, you can consult any calculus book, and get a
good picture. We will only clarify the Integration's relation
with Differentiation.
Here, Δx.(y) ≥ ΔA ≥ Δx.(y-Δy)
∴ y ≥ Hence, Integration is reverse of Differentiation! Ok, ok, I'm not as elaborate as I have been, and I'm not going to be in this case. The concept of Differentiation is another eye to calculus. A poem in it its own right and should be enjoyed separately. It can take an essay by itself. However, it is lot more obvious to a discerning mind, than Integration. But the point to whet your appetite is, that since Integration is Anti-Differentiation, which is simpler, you could discover the value of integral by retracing the path of the differentiation. Maybe, now you have to solve some problems, to make all these concepts concrete.
The aim of this article is not to train you in solving problems on limits, integration or differentiation, but only to put all these in perspective, and relate each other. Once the concepts are clear and placed, the techniques, formulas are just modalities. Of course they too require attention and appreciation, so please dip into them with all care. Learning a field is a science and art by itself. You can see the techniques involved in it, once you learn sufficiently many different fields. Thereby you learn the invariants of learning, and hence apply it for learning a completely new thing. One constantly recurring theme is to find a basic question, the answer to which forms the proof for the existence of the field itself. Getting this question is not too tough. All you need to do is to suspect the very requirement of this new field for human knowledge. Try your level best to circumvent using any of the new field's tricks and go from the first principles. You'll soon arrive at a point, where you find the real need for this new field. Of course, if you are persevering, creative and lucky enough, you might be trail blazing a new science. Even otherwise, you loose nothing, you would have learned the subject so well in this process, that you could have never done so, had a book been given to you, and asked to write an exam in it. Maybe, if the present day schooling concentrated more on learning, enjoying and researching, than on getting to know lot of facts, we might be progressing much faster. I have myself witnessed, that this knack of asking too many unconventional questions, and re-inventing sciences, not only solves a given problem, but becomes a habit and helps me solve newer problems (in hitherto unconnected domains). This is sort of a simulated, fast forward reinvention. If calculus had taken 200 years to evolve and 200 more years to mature. We might take just around 2 weeks or 2 months to re construct the whole drama, if we are willing. Of course, never hesitate to read a book on it, but also be aware not to use anything which you have not understood. In this sense,
Analytical Geometry & Calculus
This is a book by Thomas and Finney. A masterpiece. It does cover the topics that are explained here, and lot more. The only benefit of this article would be its personal nature, tone and the simple sequence. |