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- Thread starter Gale
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chroot

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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

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HallsofIvy

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[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.

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Hurkyl

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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'

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selfAdjoint

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

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krab

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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.Originally posted by Gale17

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'

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.

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these layers have infinitesimal thickness, right?

cheers,

phoenix

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Hurkyl

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

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

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

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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?

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chroot

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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.

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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.Originally posted by Gale17

whats is stokes theorem?

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.

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