Answer
$\approx 1.84 \times 10^{12} \ J$
Work Step by Step
The force of one layer is equal to:
$\ Force= \ Mass \times \ gravity = 19600 (230-\dfrac{230 \ y }{146})^2 \Delta y \ N$
Therefore, the work done against can be computed as:
$ W=\int_{0}^{146} 19600 (230-\dfrac{230 \ y }{146})^2 \Delta y \ d y\\=19600 \pi \int_{0}^{146} (52900-\dfrac{52900 y }{73}+y^2 (\dfrac{230}{146})^2) y \ d y\\ =19600 [26450 y^2 - \dfrac{52900}{219} y^3 + (\dfrac{230}{146})^2)\dfrac{y^4}{4}]_0^{146} $
By using a calculator, the required result is:
$W \approx 1.84 \times 10^{12} \ J$
