Question: For the scalar field [itex]\Phi = x^{2} + y^{2} - z^{2} -1[/itex], sketch the level surface [itex] \Phi = 0 [/itex]. (It's advised that in order to sketch the surface, [itex]\Phi[/itex] should be written in cylindrical polar coordinates, and then to use [itex]\Phi = 0[/itex] to find [itex] z [/itex] as a function of the radial coordinate [itex]\rho[/itex]).
I've done as advised, and have found:
[itex] \Phi = \rho^{2}cos^{2}(\phi) + \rho^{2}sin^{2}(\phi) - z^{2} - 1 [/itex]
and therefore
[itex] z = \sqrt{\rho^{2} - 1} [/itex]
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 [itex]\Phi[/itex] in Cartesian coordinate form as [itex] z(x,y) [/itex] i.e. [itex] z = \sqrt{x^{2} + y^{2} - 1} [/itex], 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.
I've done as advised, and have found:
[itex] \Phi = \rho^{2}cos^{2}(\phi) + \rho^{2}sin^{2}(\phi) - z^{2} - 1 [/itex]
and therefore
[itex] z = \sqrt{\rho^{2} - 1} [/itex]
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 [itex]\Phi[/itex] in Cartesian coordinate form as [itex] z(x,y) [/itex] i.e. [itex] z = \sqrt{x^{2} + y^{2} - 1} [/itex], 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.
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