91Ó°ÊÓ

given,

Find the hydrostatic force exerted on one of the long sides of a rectangular swimming pool that is 20 feet long, 12 feet wide, and 6 feet deep.

Consider that the top of the tank is at height $y=6$and the bottom of the tank is at height $y=0$.

Draw a diagram that shows a thin representative slice of the tank at some point $y_k*$ from the bottom.

Assume that the entire thin slice of the wall is at a depth of $d_k=6−y_k^{∗}$units.

The area of the representative wall slice is $A_k=240â–³y$square feet.

The water density is $\mathrm{Ó\neg }=62.4$pounds per cubic foot.

The hydrostatic force exerted by the water of weight-density $\mathrm{Ó\neg }$and depth $d$on a horizontal wall of area $A$is given by $F=\mathrm{Ó\neg }Ad$.

Substitute $A_k=240\mathrm{Δ}y,d_k=6−y_k^{∗}$and $\mathrm{Ó\neg }=62.4$in $F=\mathrm{Ó\neg }Ad$to obtained $F=62.4\left(6−y_k^{∗}\right)240\mathrm{Δ}y$.

The hydrostatic force on the entire sidewall is approximately $F=\underset{k=1}{\overset{n}{∑}} 62.4\left(6−y_k^{∗}\right)240\mathrm{Δ}y$

As role="math" localid="1649309762152" $n→\mathrm{∞},F=\underset{k=1}{\overset{n}{∑}} 62.4\left(6−y_k^{∗}\right)240\mathrm{Δ}y$becomes a definite integral.

Accumulate the slices from $y=0toy=6$in order to obtain the hydrostatic force on the entire sidewall.

$\begin{array}{r}W={∫}_0^6 62.4\left(240\right)(6−y)dy\\ =62.4\left(240\right){∫}_0^6 (6−y)dy\\ =62.4\left(240\right)\left[6{∫}_0^6 dy−{∫}_0^6 ydy\right]\end{array}$

Use the power rule to evaluate the integral $62.4\left(8\right)\left[6{∫}_0^6 dy−{∫}_0^6 ydy\right]$

role="math" localid="1649310053685" $\begin{array}{r}W=62.4\left(240\right)\left[6{∫}_0^6 dy−{∫}_0^6 ydy\right]\\ =62.4\left(240\right){\left[6y−\frac{y^2}{2}\right]}_0^6\\ =62.4\left(240\right)\left[6\left(6\right)−\frac{(6{)}^2}{2}\right]\\ =269,568\end{array}$

The hydrostatic force is $269,568$pounds.