91Ó°ÊÓ

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 bottomof 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 wall is at a depth of $d_k=6-y_k^{*}units$.

The area of the representative wall slic is $A_k=240∆y$square feet.

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

The hydrostatic force exerted by a water of weight-density $\mathrm{Ó\neg }$and depth density $d$on a horizontal line of area $A$is given by

$F=\mathrm{Ó\neg }Ad$.

Substitute , $A_k=240∆y,d_k=6-y_k^{*},\mathrm{Ó\neg }=62.4$in $F=\mathrm{Ó\neg }Ad$to obtain

$F=62.4\left(6-y_k^{*}\right)240∆y$.

The hydostatic force on the entire side wall is approximately

$F=\underset{k=1}{\overset{n}{∑}}62.4\left(6-y_k^{*}\right)240∆y$

As $n→∞$, $F=\underset{k=1}{\overset{n}{∑}}62.4\left(6-y_k^{*}\right)240∆y$becomes a definite integral.

Accumulate the slices from $y=0toy=6$in order to obtain the hydostatic force on the entire side wall.

$$\begin{gathered} W={∫}_0^{6}62.4\left(240\right)\left(6-y\right)dy \\ W=62.4\left(240\right){∫}_0^6\left(6-y\right)dy \\ W=62.4\left(240\right){\left(6y-\frac{y^2}{2}\right)}_0^6 \\ W=269,568 \end{gathered}$$.