I worked some on this earlier, and botched it a bit, so here goes again.

Question

A water tank is made by rotating f(x)=2^x-1 between [0,2] about the y-axis. The water tank is initially full, when a hole is opened at the bottom tip so that the water drains at a rate of 4 unit^3/second. How fast is the height of the water decreasing when the height has decreased to 1?

You mentioned a possible start to a solution, but did not show any work. Here is some, with a small simplification. Instead of 2^x-1, I used e^x-1, just to get rid of all those pesky ln(2)'s floating around.

Using discs, the volume of the full tank is

∫[0,e^2-1] πln^2(y+1) dy = 2π(e^2-1)

Using shells, the volume is

∫[0,2] 2πx(e^x-1) dx = 2π(e^2-1)

So, that is apparently the proper volume integral. You can change back to using 2^x-1 and use ∫[0,3] ... dy and get a more complicated answer, but the two will still agree.

Now. For the volume as a function of h, we have

V(h) = ∫[0,h] π(ln(y+1)/ln(2))^2 dy

It doesn't really matter what the explicit value of V(h) is, because we are only interested in its rate of change: dV/dh

dV/dt = dV/dh dh/dt
we know that dV/dt=-4 and we want dh/dt when h=1.

We know that dV/dh = π(ln(h+1) /ln(2))^2, so we now have

π(ln(1+1)/ln(2))^2 dh/dt = -4
π dh/dt = -4
dh/dt = -4/π units/sec

Answer 1

dV = surface area * dh

so
dh/dt = dV/dt/surface area = Q/area