## Calculus (3rd Edition)

$1891.6 \ J$
The volume of one layer is equal to: $(0.64-y^2) \pi \Delta y \mathrm{m}^{3}$ The force of one layer is equal to: $5880 \pi (0.64-y^2) \Delta y \ N$ Therefore, the work done against can be computed as: $W=\int_{0}^{0.8} 5880 \pi (0.64-y^2) \Delta y \ d y\\=5880 \pi \int_{0}^{0.8} (0.64-y^2) y \ d y\\ = 5880 \pi [0.32 y^2-\dfrac{y^4}{4}]_0^{0.8} \\=5880 \pi [\dfrac{64}{625}] \\ \approx 1891.6 \ J$