Why Calculus?

Volume of Sphere & Calculus: From a layman's point of view

S V Ramu (2000-11-20 -Version 2.0-)

Volume of Sphere

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 solid! But, as much as the Circle (The perfect figure), its nature is hard to grasp.

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 Discrete and Continuous, Finite and Infinite. And, eventually, Newton's genius crystallized all these efforts into a supreme mechanism, called Calculus.

The Concept of Limit

It was Xeno, who first expressed the enigma of infinity and hence the limit, through his famous Achilles Paradox. It goes something like this...

Once a tortoise challenged Achilles (a then famous Greek runner and Olympic champion), that he cannot beat it in a running race! All it requested for is a meter of handicap for him (i.e. It'll start a meter ahead of him in the race). The reasoning was, if he has the one-meter handicap, and if they start the race at the same time, then in the time that Achilles takes to cover that distance of one meter, the tortoise would have covered some distance (however small it may be). And, arguing in the same way, while Achilles cover this gap, it would have gone some more distance! And so on... Hence Achilles can never overtake the tortoise, as it will always be ahead due to his handicap!

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,

¹/₂ + ¹/₄ + ¹/₈ + ¹/₁₆ + ¹/₃₂ + ¹/₆₄ + ...

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?!

Consider, a one-meter stick. Break it into two. Put one piece into your pocket, and break the other half into two. Again, do the same, keep one of the resulting quarter piece in your pocket and break the remaining quarter into two again. And continue doing this ad infinitum. Will this infinite process of breaking-into-two result in a stick that is larger than what we started (i.e. One meter)? Never! That is the clue. In fact, the total length of the pieces in your pocket can never be more than One! Thus we can say, that after the infinite steps, the total size in the pocket will be exactly equal to One! Thus,

¹/₂ + ¹/₄ + ¹/₈ + ¹/₁₆ + ¹/₃₂ + ¹/₆₄ + ... = 1

Hence the proof! The simple fact is that the tortoise will be ahead of Achilles, for only a distance of One meter after the start, after which Achilles will out run it. (In reality, the tortoise cannot manage the lead for even few centimeters, as its speed is many times slower than even us).

The resolution of this paradox was the first step in comprehending the nature of infinity. From here on, the possibility of convergent infinite series was an established fact.

Sum to 'n' Squares

Limits Revisited

You all should have heard the story of Six Blind Men and the Elephant. The story has an interesting connotation for Human Comprehension. An abstract concept is like the elephant in this story, it is invisible to us; i.e. We are blind in that respect. If we cannot see a concept, how else will we know about it? Even we cannot see a concept fully, we can usually Sense some part of it at times. So our task is to Sense it in as many angle as possible and piece it together to form a tentative full picture.

What does all these mean to our understanding of Limits? The Limit is an abstract concept. It is also a technique. As we saw in the case of summation of series in the last section, a technique should be explored fully to make it general. We are going to do just that.

We saw that,

¹/₂ + ¹/₄ + ¹/₈ + ¹/₁₆ + ¹/₃₂ + ¹/₆₄ + ... (when continued forever!) = 1

This is Limits! Here, the limit is to infinity. What 'am I saying? Let me try in a different way again. We can write,

¹/₂ + ¹/₄ + ¹/₈ + ¹/₁₆ + ¹/₃₂ + ¹/₆₄ + ... (when continued forever! To infinity)

As, Limit n tends to infinity, of ¹/₂ ⁿ = 1

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 ¹/₂ ⁿ steadily decreases, as the n increases. So, let us specialize on the Limit n tends to infinity case fully.

First off, remember, we haven't said that all infinite series is convergent (sum up to a number). For example, it is can be proved (How? Do pursue sometime...!) that,

¹/₁ + ¹/₂ + ¹/₃ + ¹/₄ + ¹/₅ + ¹/₆ + ... (when continued forever) = infinity!

So, we have to be cautious. How? Let us clear some basic stuff. Which is bigger? ¹/₁₀ or ¹/₁₀₀? Of course, the one with the larger denominator is the smallest (if the numerator is same). So the decreasing order is ¹/₁₀ , ¹/₁₀₀, ¹/₁₀₀₀ .... Now, what happens if the denominator keeps increasing? The fraction keeps decreasing. But what will it be ultimately? Zero of course! (Please pursue further if not convinced). Thus we have,

Limit n tends to infinity, of ¹/_n = 0

Aha! That is something! Now we turn the question on its head. What happens if the denominator becomes smaller and smaller?

¹/₁₀₀ = .01
¹/₁₀ = .1
¹/₁ = 1
¹/_.1 = 10
¹/_.01 = 100
¹/_.001 = 1000
...

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 ¹/_n = infinity

Great! We are getting somewhere. To simplify things,

¹/_infinity = zero

And,

¹/_zero = infinity

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 n tends to infinity.

Limit n tends to infinity, of (n + 1)(2n + 3) / 5n²
= (2n² + 5n + 3) / 5n²
= 2n²/5n² + ⁵ⁿ/5n² + ³/5n²
= ²/₅ + ⁵/_5n + ³/5n²

Now, making the n as infinity, the denominator of the 2^nd and 3^rd term becomes infinity (Believe me! 5 × infinity = infinity. Too much to believe? Then test it!), hence those terms become zero! What is left is ²/₅.

Thus, Limit n tends to infinity, of (n + 1)(2n + 3) / 5n² = ²/₅

Do you see the power! We shall need these analysis and techniques in finding the Volume of Sphere.

Volume of Sphere Revisited

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 n circular disks of equal thickness. So the thickness of each disk will be R/n, where R is the radius of the sphere. Also the radius of the k-th disk can be found by the famous Pythagoras Theorem (Square of the hypotenuse equals the sum of the squares of the other two sides. How?!). Here,

R² = r² + (k . R/n)²
Thus, r² = R² – (k . R/n)²

Now, what is the Volume of the Whole Sphere?

= 2 times the volume of each hemisphere
= 2 × Sum of the Volume of all the disks (i.e. cylinders) in an hemisphere

You should understand that each disk is a cylinder, and,

Volume of a cylinder = Area of its base ^x its height = πr² . (R/n)

Thus the Volume of the Sphere is,

= 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 Sum to n Squares (that we derived earlier) formula here,

= 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] }
= 2 πR³.{ 1 – (1/n)² [(n + 1)(2n + 1)/6] }
= 2 πR³.{ 1 – [(n + 1)(2n + 1)/6n²] }
= 2 πR³.{ 1 – [(2n² + 3n + 1)/6n²] }
= 2 πR³.{ 1 – [2n²/6n²+ 3n/6n² + 1/6n²] }
= 2 πR³.{ 1 – [1/3+ 1/2n + 1/6n²] }

Well! This is the Volume of the Sphere approximately! The only reason for the approximation is the presence of n still there. After all, the Volume of the Sphere depends only on its radius R, hence there is no reason for any other variable to be present. When we started, we said that the infinite number of very small things becomes continuous itself. Thus when the number of discs become infinite, and their thickness become zero, and then the sum of the volume of all these discs will eventually become the Volume of the Sphere (V) itself!

Thus when the limit of n tend to infinity, in the above final equation, the 2^nd and the 3^rd term vanishes, thus,

V = 2πR³.{1 – 1/3} = (4/3)πR³
∴ Volume of the Sphere = ⁴/₃πR³

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.

The Newton's Genius

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 summing convergent infinite series, which happens to be translating itself as area under the curve, geometrically.

Very simply put, integral f(x)dx, between x=a and x=b (b>a) can be defined as, the area under the curve f(x), between a and b. Remember, a Definition is only an alias. No new truth is in there, just a sweet name in place rambling details. Note too, that a proper definition, clarifies immensely. Many a time, a good definition almost has the same clout as that of an Axiom, which is in fact a distillate of knowledge, that can make or break a science. These building blocks take lot of time to evolve, but once done, the problem picture becomes lot clearer.

_a∫^b f(x)dx = Area under the curve f(x), between x=a and x=b

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 = ^(b-a)/_n

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,

_a∫^b f(x)dx = ^lim _{n → ∞ k=1}∑ⁿ h.f(k.h), 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 r

Well, 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 limit of an infinite convergent series, as the area under a suitable curve, and is represented in a neat form. Yes, nothing else at all. Well, there is nothing new in this, we have already seen that, for finding its volume, we can split a continuous and smooth sphere into infinite number of discrete sections and then summing them all. The above cryptic and intimidating expression, just formally concretize our volume trick. The point is, that the volume trick is such a powerful tool, that we are going to use it for any Continuous → Discrete → Continuous round-trip magic. The proof of the pudding is in eating, so let us use this beautiful equation, to simplify the volume-of-sphere derivation.

Here h=^(R-0)/_n=^R/_n, and lim n → ∞
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, ₀∫^R f(x)dx = h. _k=1∑ⁿ f(k.h),

Here, since h acts as the height of the cylinder, f(k.h) is nothing but the area of the circle of k^th cylinder.

Now, Area of the Circle = π.radius²
where, radius² = [R² - (k.h)²] = [R² - x²], as f(x)=f(k.h)
So, area of the k^th cylinder's circle, f(x) = π.[R² - x²]
of course, height of the cylinder = h = ^R/_n
When lim n → ∞, Volume of Sphere = 2 × Volume of the Hemisphere
= 2. ₀∫^R f(x)dx = 2.^R/_n. _k=1∑ⁿ π.[R² - (k.^R/_n)²]

Isn't this wonderful! The integral form of the volume of sphere is,

Volume of Sphere = 2 ₀∫^R π.[R² - x²]dx

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 ₀∫^R π.[R² - x²]dx
Hoping that you know that, ∫ xⁿdx = n.x⁽ⁿ⁺¹⁾/(n+1)
∴ Volume of the Sphere
= 2π ₀∫^R R²dx - 2π ₀∫^R x²dx
= 2πR² [R-0] - 2π [R³/3 - 0]
= 2πR³ (1-1/3)
∴ Volume of the Sphere = ⁴/₃πR³

Integration Nirvana

The last section's goal was to relate the modern Integration techniques with the bygone summing convergent series technique, through finding the volume of a sphere. But, Integration demands its own attention too. We'll now try to bridge this gap between the old-intuitive discrete methodology to the new-elegant integral technique, by exposing their relation in various angles.

The key concept that irritates a calculus student is the concept of 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,

_a∫^b f(x)dx = ^lim _{n → ∞ k=1}∑ⁿ h.f(k.h), where h=^(b-a)/_n

_a∫^b f(x)dx = ^lim _{n → ∞ k=1}∑ⁿ h.f(a + k.h), 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 = ₀∫^x f(x)dx

Ok, what is the point in making such a definition for definite integral?

If, ∫ f(x)dx = ₀∫^x f(x)dx = F(x), say
Then, _a∫^b f(x)dx = F(b)-F(a)

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)

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)).

Integration is Anti-Differentiation

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 ≤ ^ΔA/_Δx ≤ y-Δy
∴ y ≤ lim_Δx→0 ^ΔA/_Δx ≤ y
∴ ^ΔA/_Δx = y
But we know, A = ∫y Δx

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.

Epilogue

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, Reinventing the wheel has its own usefulness for human understanding. Maybe we can only say, that do your re-inventions consciously, not as a lifetime pursuit, if you do find some new paths, do refer constantly (maybe occasionally?) to the available literature, to help you not waste too much by going too far through a well trodden path. Well, as somebody said, that a philosophy put in a nutshell stays there. The art of learning, to put it simply, is an art. How much ever you study, how much ever it can be simplified, you can never substitute a personal passion or passionate practice. Get your hands dirty, perfection is not too far (or, is it far, and it doesn't matter?)!

References

• 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.