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I am a high school student, so I know how to derive the volume $V=\dfrac{4}{3}\pi r^3$ using calculus, but I am unable to derive its surface area.

However, I notice that we can approximate the surface area of a sphere by imaging 'the shell' of the sphere as the difference between two similarly sized spheres with width infinitely small.

What I mean is that if we have two spheres, one of radius $r$ and another of radius $r+h$, where $h$ is very small, the surface area can be roughly measured by the difference in volume of these two spheres divided by its width $h$.

Therefore if $f(r)$ is the volume of a sphere with radius $r$, then the surface area should be $$\lim_{h\to0} \dfrac{f(r+h)-f(r)}{h}$$but this is exactly the derivative $f'(r)$ of the volume.

Hence we can say that the surface area of a sphere is $4\pi r^2$.

Is this proof even correct? Is it rigorous? How do I improve it and make it more convincing?

I noticed how this should similarly work for other objects, for example, the derivative of the area of a circle is its perimeter. Does this phenomenon likewise extend to other examples?

Thanks for all your help!

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    $\begingroup$ The proof is correct, rigorous and convincing. The method extends to smooth convex bodies in higher dimensions. $\endgroup$ Commented Oct 13, 2015 at 12:03
  • $\begingroup$ The only aspect that is not completely convincing is that you need to know that both the surface and volume are well-defined via certain integrals, and that "thickening" the surface leads to the volume of a corresponding larger object "without overlaps" The "convex bodies" part of @uniquesolution's comment is one way to ensure this. For things like the Koch snowflake, this relationship does not hold at all. $\endgroup$ Commented Oct 13, 2015 at 12:43
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    $\begingroup$ The proof is correct, although extending it to other objects requires some care. For instance, you might imagine that the perimeter of an ellipse with minor axis $a$ and major axis $b$ could be calculated by finding the derivative of $A(a+r, b+r)=\pi (a+r)(b+r)$ with respect to $r$ at $r=0$. This is not correct. $\endgroup$
    – mjqxxxx
    Commented Oct 13, 2015 at 12:45
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    $\begingroup$ @hypergeometric: It works for all regular polyhedra, once you realize that r is the length of the segment uniting the center of symmetry with the midpoint of each face. $\endgroup$
    – Lucian
    Commented Oct 13, 2015 at 15:29
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    $\begingroup$ Related: Why is the derivative of a circle's area its perimeter (and similarly for spheres)?. $\endgroup$ Commented Oct 14, 2015 at 12:57

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Given the multitude of different comments and given the first answer saying that your formula "hold in general", I would like to emphasize: It is NOT true that $V'(r)=S(r)$ for the Volume $V$ and Surface $S$ of general convex bodies scaled by $r$.

What is true, is that locally

$S\approx V/h$,

where locally means that we are considering a small piece of surface, small enough that it is approximately a piece of a plane, and denote by $S$ its area and by $V$ the volume of the body that you get from extending the surface along its normal direction by length $h$ (normal = orthogonal to the surface)

When you want to consider the global case, that is you want to "sum up" the local formulae, the sphere is a special case because when you enlarge the sphere then all small pieces of surface as described above are by the same normal length $h$. For other convex bodies the summing up does not work.

The problem with general convex bodies is that when you enlarge the body some of the small pieces of surface are extended more than others. For example, if you try to apply your formula to rectangles with side lengths $2r$ and $r$ (and center in the origin), you get the încorrect formula $S(r)=V'(r)=(2r^2)'=4r$. The difference to the correct formula $S(r)=6r$ comes from the fact that dividing by $h$, as you do, is "unjust" to the long sides. These long sides get extended by only $h/2$ in their normal direction when $r$ is enlarged to $r+h$.

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