Answer
$985203.5 \ J$
Work Step by Step
The volume of one layer is equal to:
$ \pi (4y-y^2) \Delta y \mathrm{m}^{3}$ .
The force required to lift the layer is equal to:
$1000 (9.8) \pi (4y-y^2) \Delta y \ N$
Therefore, the work done to fill the tank can be computed as:
$ W=9800 \pi \int_{0}^{4} (y+1) (4y-y^2) \ d y\\= 9800 \pi \int_{0}^{4} (3y^2+4y-y^3) \ d y\\ =9800 \ \pi [-\dfrac{y^4}{4}+y^3+2y^2]_0^4 \\=313600 \pi \\ \approx 985203.5 \ J$
