Question

A rectangular swimming pool 50 ft long, 25 ft wide, and 10 ft deep is filled with water to a depth of 8 ft. Use an integral to find the work required to pump all the water out over the top. (Take as the density of water δ=62.4lb/ft3.)

Answer 1

Answer:

$W= \int_{0}^8 78000 (9-h) dh$

$W = 78000 (9h - \frac{h^2}{2}) \Big|_0^8$

$W= 78000 (72-32) = 3120000$

Explanation:

For this case since w ehave a rectangular pool we can find the volume as:

$V = xyh$

We can approximate the volume with the following formula:

$V \approx 50 ft* 25 ft * \Delta h)$

The mass is defined $m= \rho V$

If we replace our approximation for the volume we got:

$m =\rho(\approx 50 ft* 25 ft * \Delta h)$

The work for this case is defined as:

Work= Density* Force * Distance

We know that the total height is 9 ft but we want to calculate the work until 8ft so then we can express the distance in terms of the heigth h like this $D= 9-h$

And then the work can be founded like this:

$W = \sum_{i=1}^n \rho (1250 \Delta h) (9-h)$

And is we convert this into integrals using $\Delta h \to 0$our integral limits are 0 and 8 and then density for the water is a property $\rho = 62.4 \frac{lb}{ft^3}$ and we have this:

$W= \int_{0}^8 78000 (9-h) dh$

$W = 78000 (9h - \frac{h^2}{2}) \Big|_0^8$

$W= 78000 (72-32) = 3120000$