Chapter 7.5, Problem 13E

$\text{(a)}$

To determine

To explain: How to approximate to force against the end of the tank by the Riemann sum.

Answer to Problem 13E

The answer is:

Explanation of Solution

Given information:

The vertical end of the tank contained water weight $62.4\text{lb}/{\text{ft}}^3$ has the given shape.

The figure is:

  

Calculation:

Let's first review the equation for a fluid force in a constant-depth surface in order to determine the force acting against the end of the tank.

  $\begin{array}{c}F=pA\\ =whA\end{array}$

Where,

  $\begin{array}{c}p=\text{ pressure }\\ A=\text{ area }\\ w=\text{ weight-density }\\ h=\text{ height }\end{array}$

Let's partition the height of the tank into k parts in order to approximate the force using the Riemann sum. The semicircle shape of the tank suggests that it has a radius of $3ft$ . Place the tank in the middle of the cartesian coordinate plane to get the first strip, which is shown below.

  

Where,

  $\begin{array}{c}2x_i=\text{ strip depth }\\ \sqrt{9-x^2}=\text{ strip length }\\ \text{Δ}x=\text{ width of the strip}\end{array}$

Now, The Riemann sum formula for the force is given below.

To generate the force approximation, substitute the value from the graph into equation $(1)$ along with the weight-density specified in the issue $(62.4lb/f{t}^3)$:

$\text{(b)}$

To determine

To find: The force as an integral.

Answer to Problem 13E

The answer: $1123.2\text{lb}$

Explanation of Solution

Given information:

The vertical end of the tank contained water weight $62.4\text{lb}/{\text{ft}}^3$ has the given shape.

The figure is:

  

Calculation:

Have a look at the graph above. The y-axis of the graph is symmetric, as can be seen. multiply by two and integrate the right side using the same strip. Put the integral together,

  

Now let,

  $\begin{array}{c}u=9-x^2\\ du=-2xdx\\ xdx=-\frac{du}{2}\end{array}$

New boundaries of integration:

Upper boundary:

  $\begin{array}{c}x=3\\ $ \to u=9-{(3)}^2$=0\end{array}$

Lower boundary:

  $\begin{array}{c}x=0\\ $ \to u=9-{(0)}^2$=9\end{array}$

Put everything to the integration:

  

Now evaluate the integral:

  

Therefore, the required force as an integral is $1123.2\text{lb}$ .