# Thread: Volume, derivative and surface area.

1. ## Volume, derivative and surface area.

Hi all ,

I recently came across an article on Archemedes about his finding the formula for the surface area of a sphere. An alternative way to find it is to consider the volume as the summation of an infinity of shells of infinitesimal thickness having the value of the surface area, calculating the derivative logically gives the surface area as the increment x+h (infinitesimal) makes the incremental volume coincide with the infinitisemally thick shell.
4/3*Pi*R^3---->4*Pi*R^2.

So I inferred that such reasoning must be valid for all volumes and surface areas but it is not.

I tried with the cube V=X^3 and derivative is 3x^2, the surface area is actually twice that = 6x^2.

Could someone explain why this reasoning only seems to work with a sphere ?

Note : I am not a math person and do not do maths as a profession, so simple terms are welcome

2. Originally Posted by Werecat
Hi all ,

...

So I inferred that such reasoning must be valid for all volumes and surface areas but it is not.

I tried with the cube V=X^3 and derivative is 3x^2, the surface area is actually twice that = 6x^2.

Could someone explain why this reasoning only seems to work with a sphere ?

Note : I am not a math person and do not do maths as a profession, so simple terms are welcome
This relation between volume and surface area (area and circumference) works if the origin of the coordinate system is the center of the solid and the solid is symmetric to the origin:

Let O(0, 0) be the center of a cube. Then the length of one edge is $a = 2x$. Thus the volume is

$V = a^3 = (2x)^3=8x^3$

Differentiate wrt x:

$V'(x)=24x^2 = 6 \cdot (2x)^2 = 6 \cdot a^2$

You get obviously the usual value of the surface of a cube.