Chapter 7.5, Problem 16E

a.

To determine

To approximate: the total force using Reiman sum.

Answer to Problem 16E

The Reiman sum form is ${\displaystyle \sum _{k=1}^{n}124.8\sqrt{2}\left(4\sqrt{y_k}-y_k^{\frac{3}{2}}\right)\Delta y}$ .

Explanation of Solution

Given information:

• The vertical end of the tank is in the triangular form of

  

• The weight of the water in the tank is $62.4\text{ lb}$ .

Formula used:

  $\text{Force = Pressure }\times \text{ Area}$

Calculation:

The equation of parabola can be found as:

  $y=a{x}^2$ , where a is any arbitrary constant.

Consider the bottom of the parabola as $\left(0,0\right)$ .

Then the rightmost point of the parabola will be: $\left(3,4.5\right)$ .

Put $\left(3,4.5\right)$ in $y=a{x}^2$ .

  $\begin{array}{c}4.5=a{\left(3\right)}^2\\ a=\frac{4.5}{9}\\ a=0.5\end{array}$

Hence the equation of parabola will be: $y=\frac{1}{2}{x}^2$ .

The equation of the height of the tank can be given as,

  $\begin{array}{c}y=\frac{x^2}{2}\\ x^2=2y\\ x=\sqrt{2y}\end{array}$

Hence, the horizontal rectangle area is, $2x\Delta y=2\sqrt{2y}\Delta y$ .

Where y is the distance from the bottom.

Now, Pressure is, $\text{Weight}\times \text{height of the water}=62.4\left(4-y\right)$ .

Hence the total force is approximately,

  ${\displaystyle \sum _{k=1}^{n}62.4\left(4-y_k\right)\left(2\sqrt{2y_k}\right)\Delta y}={\displaystyle \sum _{k=1}^{n}124.8\sqrt{2}\left(4\sqrt{y_k}-y_k^{\frac{3}{2}}\right)\Delta y}$ .

b.

To determine

To Find: the force as an integral and evaluate it.

Answer to Problem 16E

The integral is ${\displaystyle {\int }_0^{4}124.8\sqrt{2}\left(4\sqrt{y}-y^{\frac{3}{2}}\right)dy}$ and the force is 1506.08 lb.

Explanation of Solution

Given information:

The Reiman sum is previously obtained as ${\displaystyle \sum _{k=1}^{n}124.8\sqrt{2}\left(4\sqrt{y_k}-y_k^{\frac{3}{2}}\right)\Delta y}$ .

Therefore, $F\left(x\right)=124.8\sqrt{2}\left(4\sqrt{y_k}-y_k^{\frac{3}{2}}\right)$ and $a=0;b=4$ .

Formula used:

The work done of a moving object from a point to is ; where is the working force.

Calculation:

The total work done is,

  $\begin{array}{c}{\displaystyle {\int }_0^{4}124.8\sqrt{2}\left(4\sqrt{y}-y^{\frac{3}{2}}\right)dy}=124.8\sqrt{2}{\left[\frac{8}{3}{y}^{\frac{3}{2}}-\frac{2}{5}{y}^{\frac{5}{2}}\right]}_0^4\\ =124.8\sqrt{2}\left[\frac{8}{3}{\left(4\right)}^{\frac{3}{2}}-\frac{2}{5}{\left(4\right)}^{\frac{5}{2}}-0\right]\\ =1506.08\end{array}$