A rotating bowl?

Question

A bowl is made by rotating y=ax² around the y-axis. (a is a constant)
a) The bowl is filled with water to a depth h. What is the volume of water in the bowl? (answer will have a and h)
b) what is the area of the surgace of water if the bowl is filled to depth h? (answer will contain a and h)
c) Water is evaportating from the surface of the bowl @ a rate proportional to the surface area, w/ proportionality constant k. Find a differential equation satisfied by h as a functino of time, t. (find eqn for dh/dt)
d) If the water starts @ depth h0, find the time taken for all the water to disappear.

Answer 1

a) volume can be found by integration, i.e. to sum up very thin discs from y=0 to y=h.
Each thin discs is of radius x and thickness delta_y.
Thus, volume of a disc is pi x^2 delta_y.
summing many of these volume with small thickness delta_y for different values of y from 0 to h is
integrate pi x^2 w.r.t. y from 0 to h, where x^2 = y/a.

b) surface of water = area of the circle of radius x when y = h, i.e. pi (h/a), because x^2 = h/a in this case.

c) Given dV/dt = -k pi h/a, where V is found in a. Thus, implicit differentiating dV/dt will have some terms of dh/dt, and so you'll get the differential equation.

d) Solve for your differential equation for h in terms of t. Find the difference between values of t for h = h0 and h=0.


ps: by the way, the bowl is not rotating!