Chapter 21, Problem 1RE

Describe in words the objects parameterized by the equations in Exercises 1–2.

$x=2z\cos \theta$ $y=2z\sin \theta$ $z=z$ $0\le z\le 7$ $0\le \theta \le 2\pi$

To determine

To Explain: In words about the objects parameterized by the equations: -

  $\begin{array}{c}x=2z\cos \theta ,0\le z\le 7\\ y=2z\sin \theta ,0\le \theta \le 2\pi \\ z=z\end{array}$

Explanation of Solution

Given information:

Given equations,

  $\begin{array}{c}x=2z\cos \theta ,0\le z\le 7\\ y=2z\sin \theta ,0\le \theta \le 2\pi \\ z=z\end{array}$

Since,

  $\begin{array}{l}x=2z\cos \theta ,\\ y=2z\sin \theta \end{array}$

Then,

  $\begin{array}{c}{x}^2+y^2={\left(2z\cos \theta \right)}^2+{\left(2z\sin \theta \right)}^2,\\ =4z^2{\cos }^2\theta +4z^2{\sin }^2\theta \\ =4z^2\left({\cos }^2\theta +{\sin }^2\theta \right)\\ =4z^2\end{array}$

So, the equation results in the form given in below.

  $x^2+y^2=4z^2,0\le z\le 7$ .

This is the equation of cone and it can be written as:

  $\begin{array}{c}{x}^2+y^2=4z^2\\ x^2+y^2-4z^2=0\\ \frac{x^2}{4}+\frac{y^2}{4}-\frac{z^2}{1}=0\\ \frac{x^2}{2^2}+\frac{y^2}{2^2}-\frac{z^2}{1^2}=0\end{array}$

The axis of the surface corresponding to the variable with a negative coefficient. The trace in the coordinate planes parallel to the axis are intersecting lines.