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Area of a circle is $s=\pi r^2$ and perimeter is $p=2\pi r$ $$\frac{ds}{dr}=P\\\frac{d(Area)}{d(radius)}=perimeter$$also
Volume of hemisphere is $V=\frac{4}{3}\pi r^3$ and surface is $S=4\pi r^2$ $$\frac{dv}{dr}=S\\\frac{d(Volume)}{d(radius)}=surface$$ my question is : Is there a logic ? for this derivation ?
thanks in advance

KReiser
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Khosrotash
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    This works for spheres in any number of dimensions, but not more generally. – Angina Seng Aug 28 '17 at 16:13
  • It works in a few more cases, but it requires care. A cube of side $2r$ has volume $8r^3=(2r)^3$ and surface area $24r^2=6\cdot (2r)^2$. You just have to expand the "radius" of the cube, rather than the side of the cube. @LordSharktheUnknown – Thomas Andrews Aug 28 '17 at 16:23
  • Refer here for in-depth answers https://math.stackexchange.com/questions/625/why-is-the-derivative-of-a-circles-area-its-perimeter-and-similarly-for-sphere – Isaac Browne Aug 28 '17 at 16:30
  • I think it works with many bounded convex sets $U\subseteq \mathbb R^n$ with $0$ in the interior that if $f(r)=\mathrm{vol}(rU)$ then $f'(r)$ is the surface area of $U$. You might need some nice symmetry in addition. – Thomas Andrews Aug 28 '17 at 16:30
  • These relations arise because you can actually link the area to the perimeter by integration and the volume of a sphere to its surface, also by integration. – MrYouMath Aug 28 '17 at 16:30

2 Answers2

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It works as an approximation. If you start with a 2D shape and expand it by a thin layer on the outside, the new area is about perimeter times the width of the layer. Similarly if you start with a 3D volume and expand it by a thin layer, the new volume is about the surface area times the thickness of the layer. You can think of breaking up the new area/volume into small squares/blocs and counting them up.

Ross Millikan
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I think this is the intuition you're looking for

A sphere consists of an infinite number of infinitesimally thick spherical shells. We can calculate the volume by integrating (summing up) the volumes of such shells. Since the surface area of a sphere of radius r is $4\pi r^2$, the volume of a spherical shell of radius $r$ and thickness $dr$ must be $dV = 4\pi r^2 dr$
Hence the volume of a sphere must be ...

You can finish reading at this physics site.

Ethan Bolker
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