Chapter 8.5, Problem 15E

a.

To determine

Explain to approximate the force against the end of the tank by a Reimann sum.

Answer to Problem 15E

  ${\displaystyle \sum _{i=1}^{k}62.4\cdot 2\left(x_n-3\right)\left(\frac{3}{4}{x}_n\right)}dx$ approximately the force against the end of the tank by a Reimann sum.

Explanation of Solution

Given information: The vertical end of a tank containing water weighing $62.4lb/f{t}^3$ .

Shape- Triangle.

  

Calculation:

Divide your triangle into k partitions so that you have thin horizontal strips. Multiply each strip by water pressure. Add everything up.

  ${\displaystyle \sum _{i=1}^{k}62.4\cdot 2\left(x_n-3\right)\left(\frac{3}{4}{x}_n\right)}dx$

Hence, ${\displaystyle \sum _{i=1}^{k}62.4\cdot 2\left(x_n-3\right)\left(\frac{3}{4}{x}_n\right)}dx$ approximately the force against the end of the tank by a Reimann sum.

b.

To determine

To find the force as an integral and evaluate it.

Answer to Problem 15E

The required solution is 3705.

Explanation of Solution

Given information: The vertical end of a tank containing water weighing $62.4lb/f{t}^3$ .

Shape- Triangle.

  

Calculation:

As per the given problem,

Set up your integral with the formula you got.

  $\begin{array}{l}{\displaystyle {\int }_3^{8}62.4\left(y-3\right)\left(\frac{3}{4}y\right)dy}\\ =3705\end{array}$

Hence, the required solution is 3705.