Q. 18 - Calculus

Question

Set up and solve definite integrals to answer each of the following questions.

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.

Step-by-Step Solution

Answer

$269,568 pounds.$

Step 1. Given Information.

 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.

Step 2. Diagram.

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.

Step 2. Formulation.

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 $\omega =62.4$ pounds per cubic foot.

The hydrostatic force exerted by a water of weight-density $\omega$ and depth density $d$ on a horizontal line of area $A$ is given by

$F=\omega Ad$.

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

$F=62.4\left(6-y_k^{*}\right)240∆y$.

Step 4. Calculation.

The hydostatic force on the entire side wall is approximately 

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

As $n\to \infty$, $F=\sum _{k=1}^{n}62.4\left(6-y_k^{*}\right)240∆y$ becomes a definite integral.

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

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