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) When a mass with volume $dV$ is lifted to a height $h(P)$ then the work done is given by:
$W=h(P) (P) dV$
In order the get the total work done, we have $W=\iiint_C h(P) g(P) dV$; where $C$ is cone
b) From part (a), we have $W=\iiint_C h(P) g(P) dV=\int_0^{2\pi} \int_{0}^{H}\int_{0}^{R(1-\dfrac{z}{H})} (z)(200) r dr dz d\theta$
or, $=(2\pi) \int_{0}^{H}[(R(1-\dfrac{z}{H}))^2 (z) dz $
or, $=200R^2\pi \int_{0}^{H}(1-\dfrac{z}{H})^2 (z) dz $
or, $=200R^2\pi \int_{0}^{H}[z+\dfrac{z^3}{H^2}-\dfrac{2z^2}{H}]_0^H $
or, $=200R^2\pi [\dfrac{z^2}{2}+\dfrac{z^4}{4H^2}-\dfrac{2z^3}{3H}]_0^H $
or, $=200R^2\pi H^2 (\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{2}{3}) $
or, $=\dfrac{50 \pi R^2 H^2}{3}$
Plug in the given values:
Work done, $W=\dfrac{50 \pi (62000)^2 (12400)^2}{3}\approx 3.1 \times 10^{19} ft-lb$
