If water is pumped in through the bottom of the tank in the picture, how much work is done to fill the tank:

a. To a depth of 2 feet?

b. From a depth of 4 feet to a depth of 6 feet?

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- Jun 6th 2007, 02:53 AMharryamount of work
If water is pumped in through the bottom of the tank in the picture, how much work is done to fill the tank:

a. To a depth of 2 feet?

b. From a depth of 4 feet to a depth of 6 feet? - Jun 6th 2007, 03:21 AMPterid
Here's what I make of it.

- - -

The tank is conical, clearly 6 units high, and it looks like it's roughly 4 units in 'radius' at the top.

Call the radius at height y $\displaystyle R(y)$. If $\displaystyle R(6)=4$ and it runs linearly (as you'd expect), then in general

$\displaystyle R(y) = \frac{2y}{3}$.

Using conservation of energy, the amount of work done to fill the tank to a height $\displaystyle y$ (call it $\displaystyle W(y)$) is equal to the total*gravitational potential energy*of the water in the container.

The gravitational potential energy is found by*integrating*(calculus) over circular "slices" of water, at height $\displaystyle y$ and with thickness $\displaystyle dy$. (Let me know if you don't get this part; it's explained better with a diagram.)

The slice will have a volume

$\displaystyle \pi \left(R(y)\right)^2 dy$.

Its*mass*comes just from multiplying by the density of water, $\displaystyle \rho$:

$\displaystyle m = \rho \pi \left(R(y)\right)^2 dy$.

Finally, the potential energy of the slice is

$\displaystyle mgy = \rho \pi \left(R(y)\right)^2 dy \cdot gy = \rho \pi gy \left( \frac{2y}{3} \right)^2 dy = \frac{4}{9} \rho \pi g y^3$

(where I have substituted for $\displaystyle R(y)$.)

Then just integrate with respect to $\displaystyle y$.

$\displaystyle W(y) = \int_0^y \frac{4}{9} \rho \pi g y^3= \frac{4}{9} \rho \pi g \left[ \frac{y^4}{4}\right]_0^y = \frac{\rho \pi g y^4}{9}$.

The answer to part a) is $\displaystyle W(2)$.

The answer to part b) is $\displaystyle W(6) - W(4)$ (as the tank was already filled to 4 units.)