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The volume enclosed by a sphere of radius radius $r$ is $\frac{4}{3} \pi r^3$. The surface area of the same sphere is $4\pi r^2$. You may already have noticed that the volume is exactly $\frac{1}{3}r$ times the surface area. Explain why this relationship should be expected. One way is to consider a billion-faceted polyhedron that is circumscribed about a sphere of radius $r$; how are its volume and surface area related?

I completed a problem similar to this relating area and perimeter, but I can't seem to think of a way that volume and surface area are related... Does anyone have tips?? Thank you!

tmaths
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Amy
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    You can pile up surfaces of infinite circles varying by infinitesimal amount in ordered manner to get a hemisphere then joining another hemisphere on one end will form a full sphere. Calculus can be used to prove the result. – Love Invariants Aug 16 '18 at 21:53
  • More exactly, as observed in a Calculus book by Serge Lang, the area is the derivative of the volume w.r.t. the volume, just as the perimeter of a circle is the derivative of the area enclosed by the circle. – Bernard Aug 16 '18 at 22:27

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One argument is that the surface area should be the derivative of the volume. Since the volume is proportional to the radius cubed ($V= aR^3$ for some $a$), taking the derivative loses a power of $R$ and gains a factor of 3.

A similar argument can be made with the polyhedra hint. The sphere can be approximated with cones (note that in mathematical terminology, a "cone" need not have a circular base) having their apices at the center of the circle and their bases approximating the surface of the sphere. The volume of a cone is one third of the base times the height. The height is the radius. So the total volume is the sum of the areas of the bases of the cones, times the radius, divided by three. The surface area is just the sum of the areas of the bases. So $V = \frac{AR}3$.

Acccumulation
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Here is an arguement based on calculus.

If you think of inflating the sphere. Then the change in volume is surface area times the change in radius (for small changes.)

$dA = \frac {dV}{dr} dr = 4\pi r^2 dr$

Another way to think about it is a multifaceted polyhedra that encloses the sphere, with each face tangent to the sphere.

The polyhedra is then many pyrmids, each with volume equal to $\frac 13$ the base area times the radius. At the limit the surface area of the polyhedra equals the surface area of the sphere and the volume equals the volume.

Based on classical geometry, Archimedes showed that the surface area of a sphere projects onto a cone of equal radius and height equal to twice the radius.

While Cavileri showed that the volume of a sphere equals the volume of the same cylinder less the volume of a doubled cone that fits inside this cylinder.

https://en.wikipedia.org/wiki/Cavalieri%27s_principle

Doug M
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