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I'm trying to prove that area of some solid figures (for example a cube and a sphere) can be found using its volume. To say it I take the solid, for example a cube, I've got it's volume: $$ V_c = l^3 $$ now I increment $l$ by $h \rightarrow 0$ and subtract the initial volume: $(l + h)^3 - l^3$. Dividing it by $h$ I: $$ S_{c/2} = \frac{(l+h)^3-l^3}{h} = 3l^2 $$ It works because I get only half-cube shell applying this method.

The same can be done with a sphere: $$ V_s = \frac{4}{3}\pi r^3$$ $$ S_s = \frac{4}{3} \pi \frac{(r+h)^3- r^3}{h} = 4 \pi r^2$$ it also works because I get a spheric shell.

How is it possible that the infinitesimal volume of the shell is also the expression of the area? In particular I can't explain the sense of dividing by $h$ to find the area?

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up vote 2 down vote accepted

If you think of covering the outer surface with little cubes $h \times h \times h$, you will need about $\frac {A}{h^2}$ of them to do it, where $A$ is the area. The volume is then $Ah$. This works for any solid. The reason you got $3l^2$ instead of $6l^2$ for the cube is that the expanded cube has side $l+2h$, not $l+h$.

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The covering method explanation is good, it explains why I should divide by $h$. About the cube is ok, of course the area is $6l^2$, not $3l^2$ I got it because I considered the increment "forward" not in all directions. – Blex Dec 6 '12 at 22:15

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