Question

Show how to approximate the required work by a Riemann sum. Then express the work as an integral and evaluate it.

A circular swimming pool has a diameter of\(24{\rm{ft}}\), the sides are\(5{\rm{ft}}\)high, and the depth of the water is\(4{\rm{ft}}\). How much work is required to pump all of the water out over the side? (Use the fact that water weighs\(62.5\)\(lb/f{t^3}\)).

Short answer

The total work required to pump all of the water out over the side is\(108000\pi \approx 339292{\rm{ft}} - {\rm{lb}}\).

Step-by-step solution

Definition of Riemann sum and work done.

Riemann sum: A Riemann sum is a type of finite sum approximation of an integral.

Work done: The term "work done" refers to both the force given to the body and the displacement of the body.

Given parameters.

Diameter of swimming pool: \(24\)ft.

Sides of swimming pool:\(5\)ft.

Depth of swimming pool: \(4\)ft.

Appears to be a diameter.

Weight of the water: \(62.5\)lb/ \({\rm{f}}{{\rm{t}}^{\rm{3}}}\)

Calculating the force.

Let us consider the given parameters.

A horizontal "slice" of the water with thickness “\(\Delta x\)” has volume.

\(\begin{aligned}{}V = \pi {r^2} \cdot \Delta x\\ = \pi {(12)^2}\Delta x\\ = 144\pi \Delta x{\rm{f}}{{\rm{t}}^3}\end{aligned}\)

The force is obtained by multiplying by the weight per unit volume.

\(\begin{aligned}{}F = 144\pi \Delta x{\rm{f}}{{\rm{t}}^3} \cdot 62.5\frac{{lb}}{{f{t^3}}}\\ = 9000\pi \Delta x{\rm{lb}}\end{aligned}\)

Finding the total work done.

Evaluating further,

If we let\(x = \)the height from the bottom of the pool, then each slice of water has to move a distance of

\(d = 5 - x\)

Because, the side of the pool is\(5{\rm{ft}}\)high.

If\(x_i^*\)

Represents this height for the\({i^{th}}\)slice then we can write the Riemann sum

\(W = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n 9 00M\pi x_i^*\Delta x\)

To sum the total work, we integrate multiplying the force time’s distance. The slices become infinitely thin (\(\Delta x\)becomes dx). The water depth is\(4\)ft., so the limits are from\(0\)to\(4\).

\(W = \int_0^6 {(9000\pi )(5 - x)} dx\)

\(\begin{aligned}{} = 9000\pi \left( {5x - \frac{1}{2}{x^2}} \right)_0^4\\ = 9000\pi (20 - 8 - (0))\\ = 108000\pi \approx 339292{\rm{ft}} - {\rm{lb}}\end{aligned}\)

Therefore, the total work done is\( = 108000\pi \approx 339292{\rm{ft}} - {\rm{lb}}\).