## University Calculus: Early Transcendentals (3rd Edition)

$967,611$ ft-lbs
The work done is defined as follows: $W=(\pi r^2 h)(\rho)(d)$ Thus, we have $W=\int_{0}^{10}[\pi (100-y^2) dy] (56) (12-y)$ or, $W=\int_{0}^{10}[56 \pi (100-y^2) (12-y) dy$ or, $W=308,000 \pi \approx 967,611$ ft-lbs