Hi guys,
I was wondering, if you want to find the volume of a threedimensional object with a nonregular surface, do you simply take the integrals of the bounding curves and multiply by one another? Are there other steps that are necessary? I'm guessing there are shortcuts, but this seems like the intuitive answer.
Thanks!
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Thread: Math question  3d integrals

09182010, 07:36 AM #1
Math question  3d integrals

09182010, 08:10 AM #2
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Well it's not that easy. If you would multiply two integrals over an area you'ld get a 4  dimensional result.
If you want to use integrals for volume modelling, you can use it with solids of revolution. These are solids that revolve around an axis, as shown here:
Solid of revolution  Wikipedia, the free encyclopedia
Other than that you'ld basically slice an object into layers and add those up. This is best done via FiniteVolumeMethods, I am tho no expert on these. There are modells that basically follow an integrated area in zdirection for example and add the integrated areas of the slices in x;ydirection up to a total sum.

09182010, 08:39 AM #3
I see. Would you mind explaining how this:
V = ∫ A(h) dh
works? Not so much the arithmetic, but what's going on "visually" so to speak. I know A(h) is the area of the perpendicular crosssections as a function of h, but I guess the question is what's the integral doing?
Wait, I might have an idea. You have a crosssection area, and you have a derivative, and the integral pulls all those A(h) together in whatever direction the derivative is pointing at h. Is that it?

09182010, 09:14 AM #4
The integral is just summing over all the crosssections, which are varying in area A(h) with h. It would just be a general way to describe integrating the third dimension of something which has a cross sectional area that varies as you move through h. If you can represent A in terms of h somehow, it can become solvable.
Visually you could look at it in two dimensions. A(h) is a line representing the crosssectional area vs h, V is the area under it. Beyond that you can't really visualise the third dimension without more information, unless you think of something circular getting wider and narrower, or something like that.Freude, schöner Götterfunken Tochter aus Elysium, Wir betreten feuertrunken, Himmlische, dein Heiligtum! Deine Zauber binden wieder Was die Mode streng geteilt; Alle Menschen werden Brüder, Wo dein sanfter Flügel weilt.

09182010, 10:47 AM #5
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Ok here're some basics ( I am always talking about squares but I mean rectangles of course. Until the last picture where its really squares ):
If you want to find out the area A of an graph like this:
You could do the following:
You take a finite amount of very small quares and add them all up to get the area. "Finite amount of small squares" is called infinitesimal.
You see above that there's still space left between the squares and the graph. To find out the area of the first square you would say: f(a) * h
To not loose the area that is missing between the first square and the graph, you say: f(a+h)*h  f(a)*h = area square one. Or better: [f(a+h)f(a)]*h
What you get is the median of the upper square and the lower square, but the result would only be a total approximation and be very unprecise. So to get the best results, you'ld have to make the squares very very small.

Step 2: Derivation
A derivation is the graph of the tangential gradient of another graph in every point.
Here's an example:
You can move the tangent on the whole graph and you always get a new point [x;y]. So a graph that would be for example be : f(x) = x² would produce a graph for its tangential gradient that would be f'(x) = 2x .
This means translated that the graph x² has a gradient of 2x. So if the graph x² would for example show the acceleration of an engine on yaxis and time on xaxis, you can say it will take 2*x seconds to get from i.e. 25 mph to 100 mph.
Now with the derivate showing us every gradiental point in one graph, why not take the points and make lines from them down to xaxis, which serve us as squares for our problem ?
To do that tho we would need the tangential gradient of a graph / function and so we just take our function we already have an say this is a tangential gradient.
This's our initial problem. So lets squeeze the square into a line. This's best done if you let it go as near to zero width as possible. For that the mathematicans use a method called Limen. This is the latin name for the borders of the roman empire. One Limes or fortified wall was built at the Rhine, which they never crossed because the barbarian germans and teutons would have whipped their asses .
So here's an intresting correlation. While the derivation of a function is it's gradient in every point, the antiderivate of a function or the primitive would be the area the function maps between 2 points and the x  axis (see first picture above).
So as you can see here:
the green area of the red function f(x) between 0 and 2 is obviously 2 squares big. One full + 2 halfes.
And this is how you solve it via integration:
Hope this makes a bit sense. General rule of thumb is: a derivation is a graphs / functions gradient in all points and presents the function f' while a primitive represents the area a graphs maps between two points a,b and the xaxis and is represented by the notation F.

09182010, 03:55 PM #6

09182010, 05:18 PM #7
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Nono ! Tis a common mistake:
An integral has nothing to do with a point, area, volume or nspace in general. It is a function. If you calculate the area under a graph via integration, you will map the difference between 2 primitives, which is F(b)  F(a) the single primitive tho is only a function / graph.
So the triple  primitive is just another function / graph transformed from an origin.

09182010, 06:16 PM #8

09182010, 06:43 PM #9
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Ah I see yes you are right, I forgot about that approach. That way you can get the volume too

09192010, 10:50 AM #10
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