Q. 18 - Calculus

Question

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

Ans:  The hydrostatic force is$269,568$ pounds.

Step 1. Given information.

 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. 

Step 2. The objective is to calculate 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.

Step 3. Now,

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

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

The water density is $\omega =62.4$ pounds per cubic foot.

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

Substitute $A_k=240\mathrm{\Delta }y,d_k=6-y_k^{\ast }$ and $\omega =62.4$ in $F=\omega Ad$ to obtained $F=62.4\left(6-y_k^{\ast }\right)240\mathrm{\Delta }y$.

 

Step 4. Therefore, the hydrostatic force exerted on one of the long sides of the hot tub is F = 62.4 6 − y k ∗ 240 Δ y .

 The hydrostatic force on the entire sidewall is approximately $F=\sum _{k=1}^n 62.4\left(6-y_k^{\ast }\right)240\mathrm{\Delta }y$

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

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

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

Step 5. The power rule for differentiation states that ∫ x n d x = x n + 1 n + 1 , where n is a real number.

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

    $\begin{array}{r}W=62.4\left(240\right)\left[6{\int }_0^6 dy-{\int }_0^6 ydy\right]\\ =62.4\left(240\right){\left[6y-\frac{y^2}{2}\right]}_0^6\\ =62.4\left(240\right)\left[6\left(6\right)-\frac{(6{)}^2}{2}\right]\\ =269,568\end{array}$

The hydrostatic force is $269,568$ pounds.