Volume and changing water level of a hemisphere

[Taibhrigh]

I've got the problem:

Water is leaking out of the bottom of a hemispherical tank of radius 8 feet at a rate of 2 cubic feet per hour. The tank was full at a certain time. How fast is the water level changing when its height is 3 feet? Note: The volume of a segment of height h in a hemisphere of radius r is (pi)(h^2)[r-(h/3)].

Here is the work I have:

(pi)(h^2)[8-(h/3)] = V

8(pi)(h^2) - (pi)[(h^3)/3] = V

dV/dt = 16(pi)(h)(dh/dt) - (pi)(h^2)(dh/dt)

Then I substituted 3 in for h, 5 in for dh (8 - 3 ?), and 2 for dt.

dV/dt = 48(pi)(5/2) - 9(pi)(5/2)

= 306.3

But, the book said the answer was -.016 ft/h

So... Either I'm doing something horridly wrong, or I haven't carried the problem far enough, and I don't know what to do next...

 

[skeeter]

you are solving for dh/dt ... dV/dt = -2

dV/dt = 16(pi)(h)(dh/dt) - (pi)(h^2)(dh/dt)

-2 = 16(pi)(3)(dh/dt) - (pi)(9)(dh/dt)

dh/dt = -2/(39pi) = -0.016 ft/hr

 

[Taibhrigh]

Thank you!