The parametric equation of a cone $z = \sqrt{x^{2} + y^{2}}$.

Question

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 $2 \pi$?

Answer 1

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.