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?
Here's what I make of it.
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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.)