- #1

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This equation in R^3 (three dimension) defines the cylinder of radius R whose central axis is the vertical line through (a,b,0).

How does it define a cylinder?

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In summary, an equation of (x-a)^2 + (y-b)^2 = R^2 defines a circle in three dimensions, while an equation of x^2 + y^2 + z^2 = R^2 defines a shell/sphere in three dimensions.

- #1

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This equation in R^3 (three dimension) defines the cylinder of radius R whose central axis is the vertical line through (a,b,0).

How does it define a cylinder?

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- #2

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Let a = b = 0 and you should clearly see that what you have is a circle centered about the origin (and, in 3-dimensions, a cylinder as those positions satisfy the equation for any 'z')

- #3

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Pengwuino said:

Let a = b = 0 and you should clearly see that what you have is a circle centered about the origin (and, in 3-dimensions, a cylinder as those positions satisfy the equation for any 'z')

It is the fact that z isn't in the equation at all.

How does the a cylinder satisfy the equations for any z if there isn't any z? I feel like this is too hand-wavy. For all I'm concerned, the domain is simply that of a two dimensional plane, more precisely a circle of radius R; no three dimensional counterpart is included.

- #4

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Now, if you had something like [itex]x^2 + y^2 + z^2 = R^2[/itex], you do have a restriction on z as well. This is the equation of a shell/sphere in 3-dimensions. The z coordinate has become restricted to the surface of this shell.

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Pengwuino said:

Now, if you had something like [itex]x^2 + y^2 + z^2 = R^2[/itex], you do have a restriction on z as well. This is the equation of a shell/sphere in 3-dimensions. The z coordinate has become restricted to the surface of this shell.

Well that is odd, I'm used to things being very clearly defined in mathematics. Z isn't anywhere in the domain and isn't anywhere in the set. I wasted a lot of time trying to understand what is going on and then I realize that the concept was just improperly conveyed. /rant

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