Sketching the Level Surface of a Scalar Field

[Insolite]

Question: For the scalar field $\Phi = x^{2} + y^{2} - z^{2} -1$, sketch the level surface $\Phi = 0$. (It's advised that in order to sketch the surface, $\Phi$ should be written in cylindrical polar coordinates, and then to use $\Phi = 0$ to find $z$ as a function of the radial coordinate $\rho$).

I've done as advised, and have found:

$\Phi = \rho^{2}cos^{2}(\phi) + \rho^{2}sin^{2}(\phi) - z^{2} - 1$

and therefore

$z = \sqrt{\rho^{2} - 1}$

However, i don't actually know how this is supposed to enable me to construct the sketch. My instinct was to rewrite the expression for $\Phi$ in Cartesian coordinate form as $z(x,y)$ i.e. $z = \sqrt{x^{2} + y^{2} - 1}$, and to plot this using a suitable program. I think i must have a deep misunderstanding of what the scalar field expression actually represents.

Any advise would be appreciated.

 

[LCKurtz]

Question: For the scalar field $\Phi = x^{2} + y^{2} - z^{2} -1$, sketch the level surface $\Phi = 0$. (It's advised that in order to sketch the surface, $\Phi$ should be written in cylindrical polar coordinates, and then to use $\Phi = 0$ to find $z$ as a function of the radial coordinate $\rho$).

I've done as advised, and have found:

$\Phi = \rho^{2}cos^{2}(\phi) + \rho^{2}sin^{2}(\phi) - z^{2} - 1$

and therefore

$z = \sqrt{\rho^{2} - 1}$

However, i don't actually know how this is supposed to enable me to construct the sketch. My instinct was to rewrite the expression for $\Phi$ in Cartesian coordinate form as $z(x,y)$ i.e. $z = \sqrt{x^{2} + y^{2} - 1}$, and to plot this using a suitable program. I think i must have a deep misunderstanding of what the scalar field expression actually represents.

Any advise would be appreciated.

If I were plotting this, I would ignore that hint. You are asked to graph the surface $x^2+y^2-z^2=1$. I would just plot a few traces, beginning with the coordinate planes. What is the curve in the $xy$ plane ($z=0$)? The $xz$ plane ($y=0$)? The $yz$ plane ($x=0$)? If that doesn't give you an idea of the shape, add a few traces for $z = \pm c$ for a few constants.

[Edit - Added]: In both of your equations with a square root, you should have $z=\pm\sqrt{...}$. The surface is symmetric in $z$. Also, with respect to what the scalar field represents, think of $\Phi(x,y,z)$ as representing the temperature at $(x,y,z)$. The level surface $\Phi = 0$ represents the surface in space where the temperature is $0$.