Chapter 13, Problem 3P

(a)

To determine

To prove: The total force exerted by the water on the 4 ft by 6 ft sides of the aquarium is $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{n}375x_k\Delta x}$ when the value of $\Delta x$ and $x_k$ is given.

Explanation of Solution

Proof:

The total force exerted on an object is calculated by below formula,

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

At a depth of x ft, the water pressure is given,

$p\left(x\right)=62.5x{\text{ lb/ft}}^2$

Calculate the force on one of the 4 ft by 6 ft sides by divide its area into n horizontal strips of width $\Delta x$, the area of each strip is,

$\begin{array}{c}\text{Area}=\text{Length}\times \text{Width}\\ \text{=6}\Delta x\end{array}$

Substitute $62.5x_k$ for Pressure and $6\Delta x$ for Area to determine the exerted force.

$\begin{array}{c}\text{Force}=62.5x_k\cdot 6\Delta x\\ =375x_k\Delta x\end{array}$

The formula of area is,

$A=\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{n}f\left(x_k\right)}\Delta x$

Thus, the total force exerted by the water on the 4 ft by 6 ft sides of the aquarium is $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{n}375x_k\Delta x}$

(b)

To determine

The significance of $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{n}375x_k\Delta x}$.

Explanation of Solution

The limit $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{n}375x_k\Delta x}$ represents the area under the graph of $p\left(x\right)=375x$ between $x=0$ and $x=4$.

(c)

To determine

To find: The limit $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{n}375x_k\Delta x}$ to find the force exerted by the water on one of the 4 ft. by 6 ft. sides of the aquarium.

Answer to Problem 3P

The force exerted on the aquarium is 3000 ft/lb.

Explanation of Solution

Given:

From part (a), the limit $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{n}375x_k\Delta x}$ is given.

The value of $x_k$ is $\frac{4k}{n}$ and the value of $\Delta x$ is $\frac{4}{n}$.

Calculation:

The total force exerted on the aquarium,

$F=\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{n}375x_k\Delta x}$ (1)

Substitute $\frac{4}{n}$ for $\Delta x$, $\frac{4k}{n}$ for in equation (1) to find the value of the total force exerted on the aquarium,

$\begin{array}{c}F=\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{n}375\left(\frac{4k}{n}\right)\left(\frac{4}{n}\right)}\\ =\underset{n\to \infty }{\lim }\frac{6000}{n^2}{\displaystyle \sum _{k=1}^{n}k}\end{array}$

The value of ${\displaystyle \sum _{k=1}^{n}k}=\frac{n\left(n+1\right)}{2}$,

Substitute $\frac{n\left(n+1\right)}{2}$ for ${\displaystyle \sum _{k=1}^{n}k}$ in the above value of F,

$\begin{array}{c}F=\underset{n\to \infty }{\lim }\frac{6000}{n^2}\frac{n\left(n+1\right)}{2}\\ =\left(3000\underset{n\to \infty }{\lim }\frac{n^2+n}{n^2}\right)\\ =3000\underset{n\to \infty }{\lim }\left(1+\frac{1}{n}\right)\end{array}$

Substitute $\infty$ for n in above values of W because the $n=\infty$ is limit,

$\begin{array}{c}F=3000\cdot 1\\ =3000\end{array}$

Thus, the force exerted on the aquarium is 3000 ft/lb.

(d)

To determine

To find: The force exerted by the water on one of the 4 ft by 3 ft sides of the aquarium.

Answer to Problem 3P

The force exerted by the water on one of the 4 ft by 3 ft sides of the aquarium is 1500 ft/lb.

Explanation of Solution

Given:

The sides of the aquarium are 4 ft by 3 ft sides.

At a depth of x ft, the water pressure is given,

$p\left(x\right)=62.5x{\text{ lb/ft}}^2$

The area is $3\Delta x$.

Calculation:

The force exerted by the water on one of the 4 ft by 3 ft sides of the aquarium is,

$\begin{array}{c}F=\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n\left(62.5\cdot 3\right)x_k\Delta x}\\ =\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{n}187.5x_k}\Delta x\end{array}$

Substitute $\frac{4}{n}$ for $\Delta x$, $\frac{4k}{n}$ for in above equation to find the value of the total force exerted on the aquarium,

$\begin{array}{c}F=\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{n}187.5\left(\frac{4k}{n}\right)\left(\frac{4}{n}\right)}\\ =\underset{n\to \infty }{\lim }\frac{3000}{n^2}{\displaystyle \sum _{k=1}^{n}k}\end{array}$

The value of ${\displaystyle \sum _{k=1}^{n}k}=\frac{n\left(n+1\right)}{2}$,

Substitute $\frac{n\left(n+1\right)}{2}$ for ${\displaystyle \sum _{k=1}^{n}k}$ in the above value of F,

$\begin{array}{c}F=\underset{n\to \infty }{\lim }\frac{3000}{n^2}\frac{n\left(n+1\right)}{2}\\ =\left(1500\underset{n\to \infty }{\lim }\frac{n^2+n}{n^2}\right)\\ =1500\underset{n\to \infty }{\lim }\left(1+\frac{1}{n}\right)\end{array}$

Substitute $\infty$ for n in above values of W because the $n=\infty$ is limit,

$\begin{array}{c}F=1500\cdot 1\\ =1500\end{array}$

Thus, the force exerted on the aquarium is 1500 ft/lb.