Identify the surface for each of the subsequent equations.

(a) r = 5

(b) r² + z² = 100

(c) z = r

Solution

(a)  In two dimensions we are familiar with that this is a circle of radius 5.  As we are now in three dimensions and there is no z in equation this means it is permitted to vary freely.  Thus, for any specific z we will have a circle of radius 5 centered on the z-axis.

Alternatively, we will have a cylinder of radius 5 centered on the z-axis.

(b) This equation will be simple to identify one time we convert back to Cartesian coordinates.

r² + z² = 100

x² + y² + z²  = 100

Thus, this is a sphere centered at the origin along with radius 10.

(c) Once again, this one won't be too bad if we convert back to Cartesian.  For reasons that will be clear eventually, we'll first square both sides, after that convert.

z² = r²

z² = x² + y²

From the part on quadric surfaces we familiar with that this is the equation of a cone.