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If we consider the equation of a circle:

$$x^2+y^2=R^2$$

Then I propose that the volume of a sphere of radius $R$ is given by the twice the summation of the circumferences of the circles between the origin and $x=R$ along the x axis, each circle having a radius equal to the value of y at that point in x.

Since

$$y={\sqrt{R^2-x^2}}$$

I derived the formula:

$$SA = 2\int^R_0{{\sqrt{R^2-x^2}}}dx$$

However, evaluating this and using integration by substitution (using $x=R\sin(u)$ to find the integral, I obtained:

$$SA=2\pi R^2\left[\frac{\sin(2u)}{2}+\frac{u}{2}\right]^{\pi /2}_0$$

I have checked this multiple times and I can't seem to see what the problem is. If the problem is with the original proposition, please could you explain why the proposition is incorrect.

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  • $\begingroup$ Here is a good source talking about surface areas of revolutions: tutorial.math.lamar.edu/Classes/CalcII/SurfaceArea.aspx. The surface area integral is done with the fustrum, not the cylinder. I'm not sure why the cylinder doesn't work, as I find, intuitively, it should. $\endgroup$ – Kaynex Feb 10 '16 at 1:40
  • $\begingroup$ The result of that integral should be the area of a semicircle. Why would you think it would have anything to do with a sphere? $\endgroup$ – Matt Samuel Feb 10 '16 at 1:41
  • $\begingroup$ @Kaynex The way I see it: both the cylinder and the fustrum methods are our mathematical attempts to model the real world. You cannot judge them unless from real life experience. And as the way they turn out, only the fustrum method provides the correct model that fits the real world. $\endgroup$ – Vim Feb 10 '16 at 2:46
  • $\begingroup$ @Kaynex see Ted Shifrin’s answer, below, and math.stackexchange.com/questions/12906/is-value-of-pi-4 for why cylinders won’t work. $\endgroup$ – amd Feb 10 '16 at 2:54
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The key thing that's going on here is that you cannot compute the hypotenuse of a right triangle by taking one of the legs of the triangle. Consider the integral that gives arclength of a curve: You're adding up $\Delta s = \sqrt{(\Delta x)^2+(\Delta y)^2}$ when you chop the curve into pieces, not adding up $\Delta x$. When you're computing your integral, you're multiplying the length of the circle by $\Delta x$, whereas you should be multiplying it by $\Delta s$.

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  • $\begingroup$ As you travel up from the equator the sphere curves away from the "infinitesimal" cylinders, so the ratio of the area of the cylinder to the area of a slice of the sphere deviates significantly from 1. $\endgroup$ – DanielWainfleet Feb 10 '16 at 6:08
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    $\begingroup$ Sure, but the flaw is there even for a cone, which doesn't "curve." $\endgroup$ – Ted Shifrin Feb 10 '16 at 6:17
  • $\begingroup$ OK this makes sense but my issue with this is that when you calculate the volume of a sphere as opposed to the surface area, the approximation I use seems to give the correct result! $\endgroup$ – Resquiens Feb 10 '16 at 11:34
  • $\begingroup$ Yes, good point. Go back one dimension, again, and think about why adding up approximating rectangles gives the right answer for area under a curve. There, we need not use trapezoids (although trapezoids give a better numerical approximation, without limits). $\endgroup$ – Ted Shifrin Feb 10 '16 at 15:37
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One way to see your error is to note that the circumference of a circle is measured in (say) inches, and the surface area of a sphere is measured in square inches. It can’t work out that adding up inches gives you square inches. Your approach is not useless, however.

Each circle you’re using contributes an infinitesimally wide strip of the surface area - say you’re looking at the circle at $x=a$. The width of that strip supplies the second dimension that creates area from lengths.

In setting up the integral, imagine that the part of the surface area at $x=a$ is the band-like sliver of area between $x=a$ and $x=a+dx$, where $dx$ is infinitesimally small. The shape of that sliver over a tiny interval of $x$-values - in particular, its width - however, changes as $x$ varies. Specifically, the width is not the constant value $dx$ (which is what you are effectively assuming). Using $dx$ for the width happens to be a good approximation for the sliver-like bands you’re adding up near $x=a$. They’re shaped like the edge of a coin, but near $x=R$, the band is shaped very differently. It’s almost flattened into a washer shape, so the width is larger than $dx$. This observation is exactly why surface area calculations need to use $ds$ (the differential or infinitesimal change in arc) instead of $dx$. There is a helpful picture here.

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    $\begingroup$ In contrast to you first two paragraphs, he is adding up length (the perimeter of the circle) times length (dx). The third is the real problem. $\endgroup$ – Ross Millikan Feb 10 '16 at 1:50
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$$ SA = 2\int^R_0{{\sqrt{R^2-x^2}}}dx $$

is not correct when summing up thin cone surface areas. Slant length is to be considered for each thin cone slice differential portion being integrated.

$$ ds= \sqrt{dx^2+dy^2} $$

Recall slant length $l$ is involved in slant area of cone in $ \pi r l ,$ not $ \pi r H$. Why do you attempt to make any difference here?

Accordingly,

$$ SA = 2 \pi \int^R_0{{\sqrt{R^2-x^2}}}ds. $$

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In this answer I'm presenting an opinion that's different from the other answers.

The other two answers seem to be trying to mathematically explain why one is supposed to integrate $\mathrm ds$ instead of $\mathrm dx$, but in my opinion, mathematics itself is unable to account for this, because they both are mathematical hypotheses to model the real world, and cannot be proved right or wrong within mathematics.

Mathematics cannot tell whether a model is right or wrong, only empirical evidence can. Both the $\mathrm ds$ and the $\mathrm dx$ are such models. They are innocent until judged empirically. And as the way they turn out, only the former fits the real world (the lateral surface area of a cone, so to speak).

To summarise, it makes little sense to judge a mathematical model only within mathematics. Only empirical evidence can tell us which models to accept and which to discard.

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  • $\begingroup$ Can I copy this answer? This is something that I've seen many people simply don't get and then ask me something along the lines of "I applied X formula but the result is wrong" or "This is solid proof that math is wrong" . And it is annoying. If I show them this probably they'll stop annoying me. $\endgroup$ – G-man Feb 10 '16 at 5:54
  • $\begingroup$ @G-man Yes sure. Just use it as you please. $\endgroup$ – Vim Feb 10 '16 at 5:56
  • $\begingroup$ I was under the impression that the Q was about the mathematics of a mathematical model. $\endgroup$ – DanielWainfleet Feb 10 '16 at 6:10
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    $\begingroup$ I totally disagree. We can compute the length of a non-horizontal line segment wrong and tell within mathematics that the answer is wrong. $\endgroup$ – Ted Shifrin Feb 10 '16 at 6:15
  • $\begingroup$ @user254665 just as I said in my answer, I don't think the model can be mathematically justified. $\endgroup$ – Vim Feb 10 '16 at 6:16

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