amount of work

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 $R(y)$ . If $R(6)=4$ and it runs linearly (as you'd expect), then in general

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

Using conservation of energy, the amount of work done to fill the tank to a height $y$ (call it $W(y)$ ) 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 $y$ and with thickness $dy$ . (Let me know if you don't get this part; it's explained better with a diagram.)

The slice will have a volume

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

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

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

Finally, the potential energy of the slice is

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

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

Then just integrate with respect to $y$ .

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

The answer to part a) is $W(2)$ .
The answer to part b) is $W(6) - W(4)$ (as the tank was already filled to 4 units.)