volume and related rates problem.... help!?

Question

A soda glass has the shape of the surface generated by revolving the graph of y=6 x^2 on [0,1] about the y-axis. Soda is extracted from the glass through a straw at the rate of 1/2 cubic inch per second. How fast is the soda level in the glass dropping when the level is 3 inches? (Answer should be implicitly in units of inches per second. Do not put units in your answer. Also your answer should be positive, since we are asking for the rate at which the level DROPS rather than rises.)

if you could talk me through it or give me hints about it that'd be great. the thing that really throws me off is where it says it has the "shape of the surface"..?????

Answer 1

The glass is formed by the volume of revolution of the parabola
y = 6x² around the y-axis for x on [0,1].

Restate the parabola as
x = √(y/6)
Revolve the parabola around the y-axis for y on [0,6].

Volume = ∫(πx²)dy = ∫{π[√(y/6)]²}dy =
= ∫{π[y/6]}dy = (π/6)(y²/2) = πy²/12

This is the volume in terms of the height y.
____________

Given dV/dt = 1/2

Find dy/dt when y = 3.

V = πy²/12
dV/dt = πy/6

dy/dt = (dV/dt) / (dV/dy) = (1/2) / (πy/6) = 3/(πy)
dy/dt = 3/(3π) = 1/π
when y = 3

Answer 2

The soda glass is the surface generated by revolving that part of the the parabola
y = 6x^2 between the y-axis and the line x=1 about the y-axis (so the glass is 6" high, and its "lip" has a radius of 1). However, it's the volume of soda (which is changing with time) inside this surface of revolution (i.e., the glass) that is relevant to this problem.

This is of course a related rates problem. You need to express the volume of soda in the glass as a function of the height y of soda in the glass.

I'm gonna leave it at that in the hope that I've adequately explained "shape of the surface."