How to sketch a surface in a three-dimensional space?

Question

I was asked to hand sketch the surface defined by

$$x^2+y^2-z^2=1$$

How could I do that? I find it particularly hard to draw graph in three-dimension, could you give me some advice?

Answer 1

Firstly, let's take a look at the surface described by the equation $z^2 = x^2 + y^2$. Since $z^2$ is always non - negative for every $z \in \mathbb R$ and $x^2 + y^2$ is non - negative as well for every $(x,y) \in \mathbb R^2$, there is no restriction for $x,y$ (see below). If we look carefully, $x^2 + y^2= r^2$ reminds us the equation of a circle centered at $(0,0)$ with radius $r\gt 0$.

Consider the case where $z \gt 0$. The equation $z^2 = x^2 +y^2$ describes a circle centered at $(0,0)$ and radius $z$. As $z$ grows, the diameter of the circles grows.

If $(x_0,y_0,z_0)$ is a point on the surface, then due to symmetry, the point $(x_0, y_0, -z_0)$ is also a point on the surface. That means we have $2$ symmetrical branches with respect to the $x,y-$ plane.

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Now, let's take a look at the equation $z^2 = x^2 + y^2 -1$. We observe that it must hold $x^2+y^2 \ge 1$, since $z^2$ is always non - negative. That means we are to exclude all the points $(x,y) \in \mathbb R^2$, which satisfy the inequality $x^2 + y^2 \lt 1$. The concept is the same as before.

Let's fix a $ z_1=c\gt 0.$ Then, we have the equation $c^2 +1 =x^2 +y^2$, which describes a circle centered at $(0,0)$ with radius $\sqrt{c^2 +1}$. Try to do so for some fixed $z'$s. Also, notice the symmetry again.

I hope I gave you a hint and here are some images of what the surface looks like. Notice the gap in the interior!

Answer 2

Hint The equation defining the surface can be written in the form $$f(x^2 + y^2, z) = 0,$$ and so the surface is symmetric about the $z$-axis. In particular, it is the surface of revolution generated by the intersection of the surface with the half-plane $$\{(x, y, z) : x \geq 0, y = 0\}, $$ and substitution of the conditions $x \geq 0, y = 0$ in the defining equation gives has the (perhaps familiar) equation $$x^2 - z^2 = 1, \quad x \geq 0, \quad y = 0.$$

Answer 3

You know how to sketch $ y^2- x^2 =1 $ already in 2-D. It is symmetrical with respect to x and y axes, and never cuts the x-axis.

So now using $ r^2 = x^2 + y^2 $ as a rotation of $r$, the given equation can be recognized as a surface of revolution obtained by revolving either of the two hyperbolas $ y^2 -z^2=1 $ , $ x^2 -z^2=1 $ about z-axis.

The swept surface is a hyperboloid of one sheet.

Now you can continue and find the surface for two sheets for further practice.