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This question started as a minecraft question, but at the heart of it, it's my lack of mathematical understanding that's hindering me.

I'm trying to understand how one mathematical formula changes format from a "y=" format to another. I'm only dealing with 3D shapes, like spheres or cylinders... or a torus.

I understand that a volume of a Torus is enter image description here

But when written in a command parser a Torus of major radius 0.75 and minor radius 0.25 is written as

(0.75-sqrt(x^2+y^2))^2+z^2 < 0.25^2

with inputs as x,y,z.

From what I can understand, the command parser will place points or blocks on any value that is greater then what is after the < symbol, in this case its 0.25^2

The proper command parser's syntax can be found here but from what I gather, the syntax is extremely common.

I've tried to understand what exactly is occurring in the transformation from one formula format to another, but I don't understand torus formula really either. Can someone show me an example of a simpler 3d shape, like a cone or a cube?

For reference, check the bottom of this page this page under Arbitrary shapes .

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Hi, This is just image a point with coordinate $(x,y,z)$ in the torus, and the inequality gives the condition on it. The left hand side established a rectangular triangle with sides of length $z$ and $0.75-\sqrt{x^{2}+y^{2}}$ respectively. If the point is strictly in the torus then the right hand side must be less than $0.25^{2}$. You may try to draw a few graphs to have a better understanding. – Kerry Jan 17 '12 at 13:07
up vote 1 down vote accepted

This is just an addendum to Shaun's answer, to honor your request for a simple example. (For the sake of simplicity I will use 2-d shapes, but the idea is exactly the same as in your case.)

Here is an inequality describing all of the points in a disc (i.e. filled-in circle) of radius $0.25$ centered at the origin of the $x$-$y$ plane:


Here is the formula for the area of the disc:


which is the special case of the famous formula $A=\pi r^2$ when the radius is equal to $0.25$.

The first inequality actually describes the coordinates of all the points $(x,y)$ that are in the disc. All points satisfying the inequality are in the disc and all points not satisfying it are not in the disc. The second equation is just a formula for one particular geometric quantity associated to the disc, its area. Other quantities (i.e. the length of its boundary; its diameter; etc.) have other formulas.

As Shaun mentioned, there is no direct way to get from the area or volume formula to the inequality describing the points of the shape; at its heart this is because different shapes can have the same area or volume. The inequality describing the points of the shape specifies not only the exact shape but its exact location in the plane. (Or in 3-dimensional space as in your example.)

However, there is a body of techniques, namely, integral calculus, to go the other way: from the equation or inequality describing all points of a shape, to numerical information about the shape such as its area, volume, etc. But it's not a straightforward thing. The calculations can be pretty intense, both conceptually and computationally.

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each answer alone wouldn't' help, but combined its great, +1 for both. I gave accepted answer the the second because it was slightly more helpful. But Thanks to the both of you. – allindal Jan 17 '12 at 19:48

If I am reading the question correctly, you are asking how the first formula listed changes to the inequality listed 2 lines below it. If that's the question, then the answer is: it doesn't.

The formula $$ V = 2\pi^2Rr^2$$ is a function which has input $r$ and $R$, the minor and major radius, respectively, and outputs the volume $V$.

On the other hand, (0.75-sqrt(x^2+y^2))^2+z^2 < 0.25^2, is an inequality describing the relationship that all points interior to the torus (with $r = 0.25$, $R = 0.75$) must satisfy. In traditional mathematical notation and with $r$ and $R$ in place of the constants, this inequality looks like: $$\left(R-\sqrt{x^2+y^2}\right)^2+z^2 < r^2.$$

There is no direct transformation from the first to the second. However, there is a way to obtain $V = 2\pi^2Rr^2$ from the inequality using advanced integration techniques.

Hope this helps!

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