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I've always thought of $dx$ at the end of an integral as a "full stop" or something to tell me what variable I'm integrating with respect to. I looked up the derivation of the formula for volume of a sphere, and here, $dx$ is taken as an infinitesimally small change which is multiplied by the area of a disc($\pi r^2$) giving $\displaystyle V = 2\pi \int_0^r x^2 dy$ which is the sum of these infinitesimals.

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Now I'm really confused. Is it correct to think of it this way? Is there any other way to prove this result without using infinitesimals?

Part two to my question: Using this same logic of using infinitesimals, I tried to find the surface area of a sphere and looked at it as the sum of infinite rings.

$\displaystyle A =2 \int_0^r 2\pi x dy$ $=> \displaystyle A = 2 \int_0^r 2\pi \sqrt{r^2-y^2} dy$ But this is wrong. Why?

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    $\begingroup$ The concept of $dx$ as an infinitesimally small quantity comes from defining the integral as a limit of a Riemann sum. A typical Riemann sum divides a region into $n$ pieces, each with a certain width, and the area of these slices is added up, being $\sum_n f(x) \cdot dx$, where the $dx$ represents the width of each slice. The integral then comes from imaginine the width of each slice approaching zero, this sum approaches a finite value $\int f(x) dx$. $\endgroup$ – Rob Bland Jun 3 '16 at 15:02
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    $\begingroup$ The symbol $dx$ does not correspond to some typographical marker implying full stop; after all, some integrals are written with the $dx$ first, as in $\int dx\ x^2$. This is particularly useful in multiple integrals, where you want to clarify which variable goes with which limits of integration. There are but few "typographical" "full stops" in mathematics, such as QED marking the end of a proof. $\endgroup$ – David G. Stork Jun 3 '16 at 15:28
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    $\begingroup$ See math.stackexchange.com/q/12906/265466. Your surface area calculation goes wrong for similar reasons. $\endgroup$ – amd Jun 3 '16 at 18:08
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    $\begingroup$ @RobBland, your comment as fine mathematically speaking, but it may be helpful to keep in mind that historically speaking, mathematicians like Leibniz (17th century) and Euler (18th century) thought about the integral as a sum of infinitely many terms. With Riemann (19th century) the subject began to be reformulated in terms more familiar today, but infinitesimals are still an invaluable tool for developing the correct intuitions. $\endgroup$ – Mikhail Katz Jun 5 '16 at 18:17
  • $\begingroup$ From a physics perspective, $\mathrm{d}x$ is more than just punctuation as it gives the right dimensions to the expression. In your case $x^2$ has dimensions of area, so $2\pi \int_0^r x^2 \,\mathrm{d}y$ has dimensions of volume. $\endgroup$ – user332714 Jun 9 '16 at 15:51
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You have to think in a geometrical way. So the line integral is against a line element $dx$. Then the area integral is against an area element $da$, the volume integral - against a volume element $dv$ and so on. The area element is an outer product of two vectors (in the sense of geometric algebra) $da =dx \wedge dy $, the volume -- of area elemnt and a line element $dv = da \wedge dz = dx \wedge dy \wedge dz $. So the coordinate transformations will produce isometric elements - i.e. scaled by the correct Jacobian of the transformation. So in this case in polar coordinates $|da | = |dx \wedge dy | = \rho \, d \rho \, d \phi $. So that $$ A = \int \int |da | = \int_{0}^{2 \pi} \, d \phi \int_{0}^{r} \rho \, d \rho = \pi r^2 $$

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The answer is simple, for volume at least. The standard Riemann integral does not require any infinitesimals to justify. Simply bound above and below by Riemann sums that tend to the same limit. This works for the Riemann sum for volume but not for the sum you suggested for surface area.

Specifically if you take the upper unit hemisphere cut into $n$ equal slices, its volume is between $\sum_{k=1}^n π(1-(\frac{k}{n})^2) \frac1n$ and $\sum_{k=0}^{n-1} π(1-(\frac{k}{n})^2) \frac1n$, which differ by $π\frac1n$ which tends to $0$ as $n \to \infty$. Since these two are Riemann sums for $\int_0^1 π(1-r^2)\ dr$, the volume sandwiched between them must be equal to $\int_0^1 π(1-r^2)\ dr$.

This argument is based on our intuitive understanding and expectation that volume is monotonic and finitely additive (when it exists), and applied to cubes gives the cube of the side length. And it generalizes to all volumes of revolution generated by a piece-wise continuous curve, giving us the familiar integral formula. If you want any further generalization, you probably want to look up the Lebesgue integral.

For surface area, you will be unable to construct sequences that give upper and lower bounds for it but tend to the same limit as your suggested sum. That means that the associated integral cannot correspond to the surface area.

Arc length and surface area, unlike planar area and volume, however, are not so easily axiomatized, and I think from the beginning one has to define it via the integral. It is then reasonable to question whether the integral captures length meaningfully. Well sort of. The arc length integral is seems to correspond exactly to the length of a flexible but inextensible string when held taut. One could say that the molecular bonds are approximately straight, so a string is nothing more than a polygonal curve with rather short segments. Indeed that is precisely the underlying meaning of the Riemann sum corresponding to the integral, which sums up the distances between consecutive points at short intervals along the curve.

I don't know of a simple intuitive justification for the integral for surface area, since it is a rather tricky thing; the naive approach of using arbitrary triangulation mesh approximations fails to match intuition. In any case, the fact remains that your integral is not going to have the same limit as whichever surface area integral currently used to define the surface area of a surface.

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  • $\begingroup$ @xasthor: My answer should address your whole question, but if you need further explanation, let me know. $\endgroup$ – user21820 Jun 13 '16 at 2:44
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The integration in question can be performed without mentioning infinitesimals explicitly. In order to understand the subtle issues involved in the calculation, it may be helpful to think of infinitesimals as Leibniz, Euler, and Cauchy did.

The subtle question here is why the residual volume (between the volume of the sphere and the "toothed" solid obtained as the infinite union of infinitely thin disks) can be neglected, whereas in evaluating the area of the surface, for example, such an approximation is inadequate. One answer is that the error involved in the volume calculation is negligible or more precisely infinitesimal, whereas the error involved in calculating the area will not be negligible.

Think for example of diagonal line in the plane. If you approximate this by a staircase then no matter how small the individual steps, the combined length of the staircase curve will not be close to the length of the diagonal.

This phenomenon occurs not only for a diagonal line of course but for any curve, for instance the circle: if you approximate it by a staircase then length will go up dramatically even if you choose very fine steps.

Since the sphere can be obtained from the circle by rotating the circle around the $z$-axis, the same sort of thing happens for the area of the sphere.

The best source for calculus with infinitesimals is Elementary Calculus.

Here $dx$ is indeed an infinitesimal, and neither a typographical device nor a full stop.

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    $\begingroup$ How would I get this same result without using infinitesimals? $\endgroup$ – xasthor Jun 6 '16 at 3:56
  • $\begingroup$ @xasthor, you don't need to reprove foundational results that can be found in any calculus textbook. The volume of a solid can be defined as an integral by exploiting the fact that it is the region under the graph of a function of two variables. The explanation for why this works and why it only works for solids is the point of my answer. $\endgroup$ – Mikhail Katz Jun 6 '16 at 7:04
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The problem with the evaluation of the area is in the term $2\pi x\,dy$, which would be the area of an infinitesimal cylinder of radius $x$ and height $dy$. But in a slice of a sphere, the surface isn't vertical and you should instead consider an infinitesimal cone, sliced in a cone tangent to the sphere at the given $y$.

The correction factor is the ratio $$\frac{\sqrt{dx^2+dy^2}}{dy}=\frac{r}{\sqrt{r^2-y^2}}.$$ If you prefer, you need to consider an element of arc rather then the element of $y$.

Hence the corrected surface

$$2 \int_0^r 2\pi \sqrt{r^2-y^2}\frac{r}{\sqrt{r^2-y^2}} dy=4\pi r^2.$$

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  • $\begingroup$ Sorry. (1) Your answer is irrelevant to the question, which explicitly asks for no use of infinitesimals. (2) Your answer even using infinitesimals is wrong; you did not at all prove that your version tends to the same limit as the definition of surface area. Prove it and I will retract my comment. $\endgroup$ – user21820 Jun 13 '16 at 2:42
  • $\begingroup$ Did you read my comments? (1) You used infinitesimals. (2) You claimed some correction factor. Not a single shred of proof is in your answer. $\endgroup$ – user21820 Jun 13 '16 at 11:11
  • $\begingroup$ Well in my opinion you in fact didn't. He asked why his way was wrong, but you only gave another way and claimed it correct. You didn't prove yours right at all, nor did you even justify it correctly. You rightly said that the surface is not vertical, but you gave no reason whatsoever why using a cone would be meaningful, since the surface is not a conical one either! In short, your answer is of the form "S is not V, so we need to use C" but what prevents someone from objecting that "S is not C, so we need to use V"? $\endgroup$ – user21820 Jun 13 '16 at 12:04
  • $\begingroup$ So you now downvote my correct answer just because yours is wrong? Good behaviour! $\endgroup$ – user21820 Jun 13 '16 at 12:07

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