# Calculating Volume Using Triangular Slices in 3D Coordinate System

• bomba923
In summary, the conversation is discussing the process of integrating triangles formed by three continuous parametric functions over a t-interval from t=a to t=b to find the total volume. The volume can be found by computing the area of each triangle using the cross product formula and integrating over the t-interval. However, this assumes there is no overlap between the triangles, which may not always be the case and could lead to a more complex integral.
bomba923
You have three continuous parametric functions (in the 3D coordinate system),
one in the xy-plane, one in the xz-plane, and the other in the yz-plane.

Then, you connect coordinates (x,t) with (y,t) with (z,t),..forming a triangle "slice" with points from these three planar functions for any given "t" (same for each point). One specific triangle for each t-value.

Then, I integrate these triangles over a t-interval from t=a to t=b. How do I find the value of this integral ? Next, how do I find the volume integrated by these triangles in the t-interval from t=a to t=b ?

It seems you're saying we have functions f, g, h : R -> R³ such that f(t) is in the x-y plane, etc. Then, for each time t, we have a triangle T(t) whose edges are the segments f(t) to g(t), g(t) to h(t), and h(t) to f(t). (You wrote some stuff about (x,t), (y,t), (z,t) but that didn't make a whole lot of sense). So, to find the area of the triangle T(t), compute:

|1/2{[f(t) - g(t)] x [f(t) - h(t)]}|

This will give you half the area of the parallelogram with 1 arm going from f(t) to g(t), and the other going from f(t) to h(t), which is precisely the area of the triangle in question. You should know this property of the cross product and its relation to the area of a parallelogram.

Next, you compute:

$$\int _a ^b \frac{1}{2}\left |\left [f(t) - g(t)\right ] \times \left [f(t) - h(t)\right ]\right |\, dt$$

H-that's right (i should have see that);
but how would one find the volume
integrated by these triangles?
(the Total volume, no overlaps)

If these are parametric funtions wouldn't we have to consier an x1=f(t) x2=g(t) y1=h(t) y2=s(t) z1=u(t) z2=v(t)?

Uhh... the first equation I gave you (in plain text) was the equation for the area of a given triangle, and the second one (in LaTeX) was the equation for the volume. Let A(t) represent the area of the triangle at time t. Then you should be able to see that the second equation given is:

$$\int _a ^b \frac{1}{2}\left |\left [f(t) - g(t)\right ] \times \left [f(t) - h(t)\right ]\right |\, dt = \int _a ^b A(t)\, dt$$

You should be able to see easily why this should give you the correct volume. The truth is, actually, that this need not give you the correct volume, as it assumes that there is no overlap between "consecutive" triangles (you should note that the above formula essentially tells you to add the volumes of infinitely many infinitessimally thin triangular prims with cross-sectional area A(t) and thickness dt, you should be used to this type of interpretation for finding volumes of solids, and if we replace A(t) with height h(t), you should be used to this interpretation for finding areas in the plane). Actually, it assumes there is no overlap between any pair of triangles, but the condition that the functions are continuous are not enough to claim that no two triangles intersect. I can't think of a general way to do this, perhaps you are to assume that they don't intersect, or that the intersections make negligible difference to the volume, and thus the volume can be computed straightforwardly as per the formula above.

That's what I was trying to figure out--the volume generated in case I do consider possible overlaps in volume between ANY multiple triangle "slices." (well, those whose overlap can affect exact volume calculation)
That would involve more complex integral(s), not as nice as the one you have presented here

Well, one thing to consider is that any two triangles will intersect on a line, and the area of the line is zero, so counting it twice has no effect, I would imagine. Then again, imagine the lines were defined by (cos(T(t-a)), cos(T(t-a)), 0) for the line in the x-y plane, where T = 200*pi/(b - a), and similarly for the other lines. These will go back and forth over and over again, and so if you computed the volume as I've given it, you'll get 100 times the actual volume. I really don't know how you could deal with the general case. Just assume that the functions are "nice".

## What is simple volume?

Simple volume refers to the amount of space occupied by an object or substance. It is typically measured in units such as liters or cubic meters.

## How do I calculate simple volume?

To calculate simple volume, you need to know the dimensions of the object or substance. For regular shapes such as cubes or cylinders, you can use a simple formula. For irregular shapes, you can use displacement method or water displacement method.

## What is the difference between simple volume and density?

Simple volume and density are two different measurements. Simple volume refers to the amount of space something occupies, while density is the measure of how much mass is contained in a given volume of a substance. In other words, density takes into account both the volume and the mass of a substance.

## Why is simple volume important in science?

Simple volume is important in science because it helps us understand the physical properties of objects and substances. It is used to determine the amount of a substance present in a given space, and it is also used in experiments to measure changes in volume. Simple volume is also important in fields such as chemistry, physics, and engineering.

## What are some common units of measurement for simple volume?

Some common units of measurement for simple volume include liters, cubic meters, gallons, and milliliters. The choice of unit depends on the size and type of object or substance being measured.

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