# Derivatives and SA

1. Sep 12, 2003

### Gale

in my calc class, someone noticed that when you take the derivative of the formula for volume of a circle, it becomes the formula for the surface area. anyone know why? and what other shapes is this also true for?

2. Sep 12, 2003

### chroot

Staff Emeritus
It's because the volume of a sphere can be found by integrating the surface area of spherical shells from the center to the surface.

Integrate[4 Pi r^2, {r, 0, R}] = 4/3 Pi R^3

It works for any shape. An area integrated over a third dimension is a volume.

For a cube, the integral looks like:

Integrate[6 Sqrt[2] s^2, {s, 0, (Sqrt[2]/2) R}] = R^3

- Warren

3. Sep 12, 2003

### HallsofIvy

Staff Emeritus
It is also true that the derivative of the Area of a circle:
[pi]r2 (with respect to r) is the circumference (2[pi]r).

That's basically because, as Chroot said, that you can find the volume a sphere by integrating the surface area of "nested" spheres or the area of a circle by integrating the circumference of "nested" circles.

4. Sep 12, 2003

### Gale

damn well, too bad i don't know what integrating is... i'm not that far into calc, but i guess i can tell my teacher that at any rate. also, what other shapes, aside from circles. i think he said something about cones... or conics... or something, i missed it.

5. Sep 12, 2003

### Hurkyl

Staff Emeritus
Any surface that, when applying a coating of paint with uniform thickness, is (sufficiently close to) the same kind of surface.

6. Sep 12, 2003

### Gale

umm... i have no clue what that means...
and since we're on the subject of my having no clue... what is this 'nesting of spherical shells'

7. Sep 13, 2003

Staff Emeritus
Like layers of an onion. There is an outside sphere, just a thin shell, and inside that is another shell whose outside surface exactly coincides with the inside surface of the first one. And ... so on... one shell inside another till you get to the center. These shells all have the same finite thickness. And then you think of having twices as many shells, each half as thick. And ... so on ... till in the limit of the thickness going to zero you recover the volume of the solid ball .

8. Sep 13, 2003

### lethe

this result also follows from stoke s theorem, one of my favorite theorems.

9. Sep 14, 2003

### krab

Here's the connection: If you paint a sphere, then the volume of paint used is the surface area times the thickness of the paint layer. Or, another way to put it is that the surface area is the volume of paint (dV) divided by the thickness of paint (dr); S=dV/dr.

This works because for a sphere, the thickness of paint is an increment in radius. It wouldn't work for, say, a cube when its volume is epressed in terms of the length of a side: V=a^3, but S is not 3a^2. However, if you write the volume of a cube in terms of _half_ a side's length h=a/2, then V=8h^3 and S is indeed 24h^2.

10. Sep 14, 2003

### phoenixthoth

"Like layers of an onion. There is an outside sphere, just a thin shell, and inside that is another shell whose outside surface exactly coincides with the inside surface of the first one. And ... so on... one shell inside another till you get to the center. These shells all have the same finite thickness. And then you think of having twices as many shells, each half as thick. And ... so on ... till in the limit of the thickness going to zero you recover the volume of the solid ball ."

these layers have infinitesimal thickness, right?

cheers,
phoenix

11. Sep 14, 2003

### Hurkyl

Staff Emeritus
An example of something with which this will not work is in order...

ellipses and other conic sections are good example; if you paint an ellipse (with an even thickness), the result is not an ellipse.

12. Sep 14, 2003

Damn you all, thats all I can ever think about when I'm painting!

13. Sep 14, 2003

### Gale

thanks, i understand now i think. i'll double check in class tomorrow. whats is stokes theorem?

14. Sep 15, 2003

### Gale

ok, i didn't understand your answers well enough to explain to my teacher, who by the way, is a wicked pain in the ass, and won't just tell me. so, could some one either explain this a different way or tell me where i could find out or something.

also: i just realized i worded my first question wrong, i said circle, i meant sphere. when you take the derative from the formula for the volume of a shere, you get the formula for it's surface area. i need to know why this is, and what shapes it holds true for, and also why?

15. Sep 15, 2003

### chroot

Staff Emeritus
Since a circle doesn't have a volume, I automatically assumed you actually meant sphere. All our responses are appropriate to the sphere.

Did you not read the responses? Or did you just not understand them?

Let's say you start with a tiny little ball bearing -- one so small that its volume is negligible.

You then paint a very thin layer of paint around the ball bearing. Then you add another layer, and another layer, and so on, until you have a baseball sized sphere. The volume is made up of many layered spherical shells of paint. You can imagine that the thickness of each of the layers is so small that they are essentially two-dimensional surfaces.

The volume of the sphere can be easily imagined to be composed of an infinite number of infinitely thin layers of paint. The surface area of each layer is 4 pi r^2. The integral (sum) of the layers is a volume, 4/3 pi R^3. (r is an internal radius, R is the radius of the outermost surface.)

If the integral of 4 pi r^2 (dr) from 0 to R (from the center to the outside of the sphere) is 4/3 pi R^3, then the converse is also true: the derivative of 4/3 pi R^3 is 4 pi r^2.

As has been mentioned, this method works for any volumes that can be built out of such layers of paint.

- Warren

16. Sep 15, 2003

### Gale

ah thanks, i was only sort of half understanding your answers. i read them, and understood what they said, but not how they related to the math i guess. would you believe that the little peice of understanding i kept missing was that the "layers of paint" were '2 dimensional' or individual sort of surface areas. i get it now though, i was just dense.

17. Sep 16, 2003

### lethe

stokes theorem says that for any space, of any dimension, the integral of some object on the boundary of that object equals the integral of the derivative of that object.

this breaks down into many familiar special cases, the fundamental theorem of calculus, the divergence theorem, the curl theorem for line integrals, the gradient theorem, and many higher dimensional generalizations.

the boundary of a filled ball is the sphere, so stokes theorem makes this relationship trivial to prove, along with many generalizations of it.