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The given equation is - $z = \sqrt{x^{2} + y^{2}} , 0 \le z \le 1$

Let $x = r \cos t$, $y = r \sin t$ and $z = r$; where $0 \le r \le 1$ and $0 \le t \le 2 \pi$.

Since $z$ is taken from $0$ to $1$ , so $r$ is also taken from $0$ to $1$

I didn't understand why $t$ is in between $0$ and $2 \pi$?

Nosrati
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Mathaddict
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  • If you wrap a piece of paper (with negligible thickness) into a cone, you can make it more than one turns, then $t$ could be beyond $[0,2\pi)$. You could illustrate the concept of isometric deformation with different domains in angle. See also parametrization of cone with geodesics in my post here. – Ng Chung Tak Dec 19 '21 at 02:19

1 Answers1

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At any given value of $r$ you get a value of $z,$ the height of the cone.

But a height and a value $r$ by themselves don't give you a cone. To have a cone you need a circular base. It will also have a circular cross section in any plane parallel to the base between the base and the apex.

Now review parametric formulas for a circle and see anything matches part of what you see here.

David K
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  • In the parametric formula of circle in $x-y$ plane, $x = r cos t$ and $y = r sin t$ . Where $t$ is the angle between $x$ axis and the position vector of point $(x,y)$. To make a complete circle, $t$ moves from 0 to $2\pi$. So, in the formula of base of cone, which is a circle, $t$ should also move from 0 to $2\pi$ to make the comlete base. Is this okay? – Mathaddict Dec 11 '18 at 06:11
  • That is the right idea. – David K Dec 11 '18 at 12:15