Question

A water tank is in the shape of a right circular cone with height 18 ft and radius 12 ft at the top. If it is filled with water to a depth of 15 ft, find the work done in pumping all of the water over the top of the tank.

Answer 1

Answer:

210,600πft-lb

Explanation:

Force is a function F(x) of position x then in moving from

x = a to x= b

Work done = $\int\limits^b_a {Fx} \, dx$

Consider a water tank conical in shape

we will make small horizontal section of the water at depth h and thickness dh and also assume radius at depth h is w

we will have ,

$\frac{w}{12} = \frac{(18-h)}{18} \\w = \frac{2}{3} (18-h)a$

weight of slice under construction

weight = volume × density × gravitational constant

$weight = \pi \times w^2 \times dh \times 62.4\\= (62.4\pi w^2dh)lb$

Now we can find work done

$W = \int\limits \, dw$

$\int\limits^{18}_{3} {(62.4\pi } \, dx w^2dh)h\\= 62\pi \frac{4}{9} \int\limits^{18}_{3} {(18-h)^2hdh} \,$

= $62.4\pi \times\frac{4}{9} (\frac{324}{2} h^2-\frac{36}{3} + \frac{h^4}{4})^1^8_3$

= 210,600πft-lb