Answer
$1.18 \times 10^8 \ J$
Work Step by Step
The volume of one layer is equal to:
$ \pi (100-y^2) \Delta y \ \mathrm{m}^{3}$.
The force of one layer is equal to:
$9800 \pi (100-y^2) \Delta y \ N $
Therefore, the work done can be computed as:
$ W=\int_{0}^{10} 9800 \pi (100-y^2) (y+2) \ d y\\=313600 [ \dfrac{(y+1)^2}{2}]_0^5 \\ = 9800 \pi \int_0^{10} 100y +200 -y^3 -2y^2 \ dy \\ =9800 \pi (5000+2000-2500-\dfrac{2000}{3}) \\ \approx 1.18 \times 10^8 \ J$
