Answer
$1.155 \times 10^7 \ J $
Work Step by Step
The force of one layer is equal to:
$\ Force= \ Mass \times \ gravity = 9800 \pi (\dfrac{y}{2})^2 \Delta y \ N$
Therefore, the work done can be computed as:
$ W=\int_{0}^{10} 9800 \pi (\dfrac{y}{2})^2 \Delta y \ d y\\=\dfrac{9800 \pi }{4} \int_0^{10} (12y^2-y^3) \ dy \\=\dfrac{9800 \pi }{4} [4y^3-\dfrac{y^4}{4}]_0^{10}\\=\dfrac{(9800) (\pi) (1500)}{4} \\ \approx 1.155 \times 10^7 \ J $
