Once the revolution is performed, there is no single value for {x,y}, rather there is a circle in space for each value of t. So it makes sense to not use revolution, but rather describe the surface more directly.

f[t_, theta_] := {(t^4 - t^7) Cos[theta], (t^4 - t^7) Sin[theta], t};

ParametricPlot3D[f[t, theta], {t, 0, 1}, {theta, 0, 2 Pi}]

Now we can choose any point on the surface by choosing t and theta, or plot the points associated with a single value of t:

ParametricPlot3D[f[.5, theta], {theta, 0, 2 Pi}]

Hi Greg,

It wasn't really bypassed! You were quite on the right track. A line segment from (0,0,t) to a point on the surface for that value of t is indeed perpendicular to the t axis. (Notice I put t on the z-axis, but that is an arbitrary choice.) And we do have the length of that segment equal to the f(t) that gives the radius of the section. But around the circular rim, for any value of f(t), the x and y values change in order to generate the circle. So we have a circle radius f(t) and as we sweep around the circle we have x and y given by f(t)cos(theta) and f(t)sin(theta). And note that for these legs of your right triangle we have the hypotenuse^2 = x^2 + y^2, which is f(t)^2*(cos(theta)^2 + sin(theta)^2). But cos^2+sin^2 = 1. So we are back to the radius = f(t).

So your right triangle observation is embedded in this solution!

Kind regards,

David