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Consider a circle with radius $R$ centered in the origin of plane $xy$. Then, consider all the inscribed circles with radius $x$ (or $y$). The length of each of the inner circles is $2\pi x$.

If I want to calculate the area of the external circle, then I can "sum" all the lengths of the inner circles as follows:

$$A = \int_{0}^R 2\pi x dx = \pi R^2.$$

In my opinion, this is a rather intuitive way to explain how integral works when we have a radial symmetry.

Another approach is the following. Cut vertically the circle and sum up twice all the lengths of the segments you obtain in this process (the segments are parallel to axis $y$ and intersect the circle in two points). Each segment has length equal to $2\sqrt{R^2-x^2}$. Then, the area is:

$$A = 2\int_0^R 2\sqrt{R^2-x^2} dx = \pi R^2.$$

Similarly, consider a sphere of radius $R$ centered in the origin of the space $xyz$. I can also consider all the inscribed spheres with radius $x$ (or $y$ or $z$), which surfaces are equal to $4\pi x^2.$

The volume of the outer sphere is given by the sum of all the surfaces of internal spheres, namely:

$$V = \int_{0}^R 4\pi x^2 dx = \frac{4}{3}\pi R^3.$$

I can find the same results by summing up twice all the areas of all the slices I can obtain from the sphere cutting along all the planes which are parallel to $yz$. In this case, the radii of the slices are equal to $\sqrt{R^2-x^2}$, and their areas are equal to $\pi (R^2-x^2)$. Then:

$$V = 2 \int_0^R \pi(R^2-x^2)dx = \frac{4}{3}\pi R^3.$$

Anyway, when I want to use a similar approach in order to find the surface of sphere, I fail. Here the intuition says me that I must sum up twice all the length of all the slices as before. These lengths are equal to $2\pi \sqrt{R^2-x^2}$, and the integral to solve is:

$$S = 2\int_0^R 2\pi \sqrt{R^2-x^2} dx = \pi^2 R^2 \neq 4 \pi R^2.$$

Why does my intuition fail here?

the_candyman
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  • I'm not sure what's going wrong at a glance, but I can say that you can get surface areas of radially symmetric objects, say $A(r)$, as $\frac{d}{dr} V(r)$. – Ian Apr 05 '17 at 20:16
  • @Ian I agree with you. Anyway I'm looking for an intuitive and "constructive" way to explain how to use integration to find geometrical entities, like areas and volumes. – the_candyman Apr 05 '17 at 20:24

1 Answers1

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Implicit to your slicing argument is an assertion about convergence of geometric quantities: you are arguing that if $V_1, V_2,\ldots V_n$ are a sequence of shapes, that converge in some sense to a limit shape $V$,and $f(V)$ is a function you want to evaluate about the shape (its volume, surface area, etc), then $$\lim_{i\to \infty}\ f(V_i) = f(V).$$

Unfortunately this property is not at all guaranteed, and its validity depends on both $f$ and the notion of convergence. For $f$ measuring volume, and for reasonably well-behaved shapes $V_i$, it suffices for the boundary of the $V_i$ to converge pointwise to the boundary of $V$. For surface area, you also need that the normals converge; the Schwarz lantern is the classic counterexample: you can construct a sequence of origami cylinders $V_i$ that converge uniformly to a smooth cylinder of unit radius and height, yet whose surface area converges to any desired value greater than $2\pi$.

A more pedestrian example of how intuitive slicing arguments can lead you astray:

enter image description here

user7530
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  • Very good argument (+1). As far as I can understand, here the limit fails since we are saying that a smooth curve (the circle) is the limit of a sequence of non-smooth curves, right? How does this relate to the example in my question? – the_candyman Apr 20 '17 at 21:24
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    @the_candyman it's more accurate to say that the normal vector to the curve is not converging to the normal vector of the circle, and this also is the reason your example fails. – user7530 Apr 20 '17 at 23:05
  • all right! thanks a lot for your answer! – the_candyman Apr 20 '17 at 23:30
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    Are you using "uniform convergence" in a different sense from https://math.stackexchange.com/a/1878310/472818 ? That answer says the Schwarz lantern doesn't converge uniformly. – mr_e_man Sep 12 '19 at 03:49
  • @mr_e_man As far as I know, the Schwarz lantern converges uniformly to the cylinder (assuming your refinement scheme for the lantern monotonically increases the number of vertices $m$ around each slice). – user7530 Sep 12 '19 at 05:12
  • @mr_e_man compare to a similar example in 1D on the unit interval, $f_{n,m}(x) = (x, \frac{1}{m} \sin(nx))$, which clearly converges uniformly to the unit interval but whose arclength does not necessarily converge to 1. – user7530 Sep 12 '19 at 05:14