Answer
a) $W=\iiint_C h(P) g(P) dV$; wher $C$ is the cone.
b) $\approx 3.1 \times 10^{19} ft-lb$
Work Step by Step
a) The work done can be calculated when a mass with volume $dV$ is lifted to a height $h(P)$ as:
$W=h(P) (P) dV$
Total work done, $W=\iiint_C h(P) g(P) dV$; where $C$ is the cone
(b) Refer part (a): $W=\iiint_C h(P) g(P) dV$
or, $W=\int_0^{2\pi} \int_{0}^{H}\int_{0}^{R(1-\dfrac{z}{H})} z \cdot (200 r dr dz d\theta)=2\pi \int_{0}^{H}[(R(1-\dfrac{z}{H}))^2 z dz =(200R^2\pi) \int_{0}^{H}[z+\dfrac{z^3}{H^2}-2\dfrac{z^2}{H}]_0^H $
and
$(200R^2\pi) [\dfrac{z^2}{2}+\dfrac{1}{4H^2}z^4-\dfrac{2}{3H}z^3]_0^H=(200R^2\pi H^2) \times (\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{2}{3}) $
Thus, $W=\dfrac{50 \pi R^2 H^2}{3}$
We need to plug the given value.
Hence, we get
$W=\dfrac{50 \pi (62000)^2 (12400)^2}{3}\approx 3.1 \times 10^{19}$ ft-lb
