Describe And Sketch The Surface.

Question

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

Consider the example where $z \gt 0$. The equation $z^ii = ten^2 +y^2$ describes a circumvolve 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 nosotros have $two$ symmetrical branches with respect to the $x,y-$ aeroplane.

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

Let's fix a $ z_1=c\gt 0.$ And so, nosotros accept the equation $c^2 +one =10^2 +y^ii$, which describes a circumvolve centered at $(0,0)$ with radius $\sqrt{c^2 +one}$. Try to exercise so for some fixed $z'$s. Also, notice the symmetry once again.

I promise I gave you lot a hint and hither are some images of what the surface looks like. Observe the gap in the interior!

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