Chapter 8.5, Problem 4QQ

a.

To determine

To find: The Riemann sum that approximates the force exerted on the entire front of the tank.

Answer to Problem 4QQ

The Riemann sum is ${\displaystyle \sum _{i=1}^{n}62.4h\cdot }2\Delta h$ .

Explanation of Solution

Given:

The breadth and height of the tank is $2ft,1.5ft$ respectively. The pressure is given by the equation $P=64.2h$ .

  

Calculation:

The Riemann sum that approximates the force exerted on the entire front of the tank can be calculated by taking the expression for pressure, and evaluating the sum over the ranges.

We have to divide $h$ into $n$ parts and evaluate.

Thus,

  ${\displaystyle \sum _{i=1}^{n}62.4h\cdot }2\Delta h$ .

Therefore, the Riemann sum is ${\displaystyle \sum _{i=1}^{n}62.4h\cdot }2\Delta h$ .

b.

To determine

To find: The definite integral that gives the force exerted on the front of the tank by using Riemann sum.

Answer to Problem 4QQ

The definite integral that gives the force exerted on the front of the tank is $140.4\text{lb}$ .

Explanation of Solution

Given:

The breadth and height of the tank is $2ft,1.5ft$ respectively. The pressure is given by the equation $P=64.2h$ .

Calculation:

To find the definite integral that gives the force exerted on the front of the tank we have to convert The Riemann integral into an integral by taking limit $n\to \infty$ .

Hence,

  $\begin{array}{c}\underset{0}{\overset{15}{{\displaystyle \int }}}62.4h\cdot 2dh\\ =140.4lb\end{array}$

Therefore, the definite integral that gives the force exerted on the front of the tank is $140.4\text{lb}$ .

c.

To determine

To find: The total force exerted on the front the tank if the front are semicircles with diameter two feet.

Answer to Problem 4QQ

The total force is $41.6\text{lb}$ .

Explanation of Solution

Given:

The information is given above.

Calculation:

Here, the equation of the semicircle with respect to height is given by,

  ${\left(\frac{x}{2}\right)}^2+h^2=1$ .

Now, from the part b our integral becomes,

  $\begin{array}{c}\underset{0}{\overset{15}{{\displaystyle \int }}}62.4h\cdot 2dh=\underset{0}{\overset{1.5}{{\displaystyle \int }}}62.4h\cdot 2\sqrt{1-h^2}dh\\ =41.6\text{lb}\end{array}$

Therefore, the total force is $41.6\text{lb}$ .