Chapter 8.5, Problem 16E

a.

To determine

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

Answer to Problem 16E

  ${\displaystyle \sum _{i=1}^{k}62.4\cdot 2\cdot \sqrt{2y_i}\cdot \left(4-y_i\right)}dx$ 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- Parabola.

  

Calculation:

First we need to parameterize the parabola. If we look at the right half, and treat the base as the origin, note that when $x=3,y=4.5$ . This tell us that $y=0.5x^2$ . The Reimann sum approximation to this given by diving the height into k partitions and evaluating ${\displaystyle \sum _{i=1}^{k}62.4\cdot 2\cdot \sqrt{2y_i}\cdot \left(4-y_i\right)}dx$ .

Hence, ${\displaystyle \sum _{i=1}^{k}62.4\cdot 2\cdot \sqrt{2y_i}\cdot \left(4-y_i\right)}dx$ 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 16E

The required solution is 1506.08.

Explanation of Solution

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

Shape- Parabola.

  

Calculation:

As per the given problem,

Which by taking the limit as $k\to \infty$ , gives us the integral,

  $\begin{array}{l}{\displaystyle {\int }_0^{4}62.4\cdot 2\cdot \sqrt{2y}\cdot \left(4-y\right)dy}\\ =1506.08\end{array}$

Hence, the required solution is 1506.08.