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 having volume $dV$ is lifted upto a height of $h(P)$, then the work done $W$ becomes $W=h(P) (P) dV$
Total work done: $W=\iiint_C h(P) g(P) dV$
(b) $W=\iiint_C h(P) g(P) dV=\int_0^{2\pi} \int_{0}^{H}\int_{0}^{R(1-\dfrac{z}{H})}\times (z) \times (200) \times r dr dz d\theta$
$=(2\pi) \times \int_{0}^{H}[(R(1-\dfrac{z}{H}))^2 \times (z) dz $
$=200R^2\pi \times \int_{0}^{H}(1-\dfrac{z}{H})^2 \times (z) dz $
$=200R^2\pi \times \int_{0}^{H}[z+\dfrac{z^3}{H^2}-\dfrac{2z^2}{H}]_0^H $
$=200R^2\pi \times [\dfrac{z^2}{2}+\dfrac{z^4}{4H^2}-\dfrac{2z^3}{3H}]_0^H $
$=200 \times (R^2\pi) \times H^2 (\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{2}{3}) $
Work done, $W=\dfrac{50 \pi R^2 H^2}{3}$
Now, substitute the given values in the above expression.
Hence, $W=\dfrac{50 \pi \times (62000)^2 \times (12400)^2}{3}\approx 3.1 \times 10^{19} ft-lb$
