Chapter 7.5, Problem 15E

a.

To determine

To approximate: the total force using Reiman sum.

Answer to Problem 15E

The Reiman sum form is ${\displaystyle \sum _{k=1}^{n}46.8\left(y_k^2-3y_k\right)\Delta y}$ .

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 lb.

Formula used:

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

Calculation:

Let x be the height of the water

The equation of the height of the water can be calculated by property of similar triangles as:

  $\begin{array}{c}\frac{x}{y}=\frac{3}{8}\\ x=\frac{3}{8}y\end{array}$

Hence, the horizontal rectangle area is,

  $2x\Delta y=\frac{3}{4}y\Delta y$

Where y is the distance from the top of the triangle.

Now, Pressure is,

  $\text{Weight}\times \text{Height of water}=62.4\left(y-33\right)$ .

Hence the total force is approximately,

  ${\displaystyle \sum _{k=1}^{n}62.4\left(y_k-3\right)\left(\frac{3}{4}{y}_k\right)\Delta y}={\displaystyle \sum _{k=1}^{n}46.8\left(y_k^2-3y_k\right)\Delta y}$ .

The Reiman sum form is ${\displaystyle \sum _{k=1}^{n}46.8\left(y_k^2-3y_k\right)\Delta y}$ .

b.

To determine

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

Answer to Problem 15E

The integral is ${\displaystyle {\int }_3^{8}46.8\left(y^2-3y\right)dy}$ and the force is 3705 lb.

Explanation of Solution

Given information:

The Reiman sum is previously obtained as ${\displaystyle \sum _{k=1}^{n}46.8\left(y_k^2-3y_k\right)\Delta y}$ .

Therefore, $F\left(x\right)=46.8\left(y^2-3y\right)$and $a=3;b=8$ .

Formula used:

The work to move an object from $x=a$ to $x=b$ is: $W={\displaystyle \underset{a}{\overset{b}{\int }}F(x)dx}$ , where $F\left(x\right)$ is the working force.

Calculation:

The total work done is,

  $\begin{array}{c}{\displaystyle {\int }_3^{8}46.8\left(y^2-3y\right)dy}=46.8{\displaystyle {\int }_3^8\left(y^2-3y\right)dy}\\ =46.8{\left(\frac{y^3}{3}-3\left(\frac{y^2}{2}\right)\right)}_3^8\\ =46.8\left(\frac{{\left(8\right)}^3}{3}-3\left(\frac{{\left(8\right)}^2}{2}\right)-\frac{{\left(3\right)}^3}{3}+3\left(\frac{{\left(3\right)}^2}{2}\right)\right)\\ =3705\text{lb}\end{array}$

The integral is ${\displaystyle {\int }_0^{8}46.8\left(y^2-3y\right)dy}$ and the force is 3705 lb.