Chapter 6.1, Problem 23E

(a)

To determine

The volume of water in the pool using the left end point values.

Answer to Problem 23E

The volume of water is $15,465{\text{ft}}^3$ .

Explanation of Solution

Given information:

Volume of Water in a Swimming Pool A rectangular swimming pool is 30 ft wide and 50 ft long.

  

Formula used:

Volume of water is given by

  $V=xbh$

Calculation:

Volume of water is given by

  $V=xbh$

Where x is length, b is width, h is depth.

Since depth is varying according to x. for LRAM, take the value of h at left end point of subinterval. Take value of x at 0, 5....45

  $\begin{array}{l}V=5b{\displaystyle \sum _{x=0}^{45}h\left(x\right)}\\ =30\left[5\left(6+8.2+9.1+9.9+10.5+11.0+11.5+11.9+12.3+12.7\right)\right]\\ V=15,465{\text{ft}}^3\end{array}$

Conclusion:

The volume of water is $15,465{\text{ft}}^3$

(b)

To determine

The volume of water in the pool using the left end point values.

Answer to Problem 23E

The volume of the water is $16,51{\text{ft}}^3$ .

Explanation of Solution

Given information:

Volume of Water in a Swimming Pool A rectangular swimming pool is 30 ft wide and 50 ft long.

  

Formula used:

Volume of water is given by

  $V=xbh$

Calculation:

Volume of water is given by

  $V=xbh$

Where x is length, b is width, h is depth.

Since depth is varying according to X. for RRAM, take the value of h at right end point of subinterval, Take value of X at 5, 10...50.

  $\begin{array}{l}V=5b{\displaystyle \sum _{x=10}^{50}h\left(x\right)}\\ =30\left[5\left(8.2+9.1+9.9+10.5+10.5+11.0+11.5+11.9+12.3+12.7+13.0\right)\right]\\ V=16,51{\text{ft}}^3\end{array}$

Conclusion:

The volume of the water is $16,51{\text{ft}}^3$ .