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We know the area of the circle is $$A(r)=\pi r^2$$Differentiate it with respect to $r$. $$A'(r)=2\pi r$$ which turns out to be the circumference of the circle.

A similar behaviour is observed with the volume of sphere $$V(r)=\frac{4}{3}\pi r^3$$ Differentiate with respect to $r$.$$V'(r)=4\pi r^2$$ which coincidently turns out to be the surface area of the volume.

Is this result purely coincidental?

The property is also found in cubes (in some fashion)

$$S(r) = 6r^2 = 2\frac{dr^3}{dr} = 2V'(r)$$ Is there something special about these shapes which can be modeled by the function$$V'(r)= kS(r)$$

mathnoob123
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    No. In the 3-dimensional case, you can imagine a ball as being made up of an infinite sequences of infinitesimal spherical shells with radius varying between $0$ and $r$. So the volume is obtained by integrating $4\pi x^{2}$ from $x=0$ to $x=r$. – preferred_anon Mar 23 '18 at 17:36
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    If you increase the radius $r$ of a circle by a tiny bit, the change in the area of the circle is in some sense the sliver given by the circumference of the circle (times a tiny thickness). Similarly, a tiny increase in the radius of a sphere increases the volume by the thin shell that is the surface area of the sphere (times a tiny thickness). – angryavian Mar 23 '18 at 17:37
  • On a related note it's also worth noticing you can't get the surface area by integrating the circumference between $0$ and $\pi$. Notionally this is because the equator moves faster than the pole. – it's a hire car baby Mar 23 '18 at 17:49

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Note that area of the circle $A$ and volume $V$ of the sphere can be obtained as

  • $A(r)=2\pi\int_0^r xdx$

  • $V(r)=4\pi\int_0^r x^2dx$

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user
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