Chapter 7.1, Problem 34E

(a)

To determine

The area of the strip is .

Answer to Problem 34E

The area of the strip is $(2\prod r)\Delta r{\text{ in}}^2$

Explanation of Solution

Given:

The radius of cylindrical pipe is $3\text{ inches}$ and the speed at which oil flows at a distance r from the center is $8(10-r^2)\text{ inches per}\mathrm{second}$ .

Calculation:

Calculate the area of contact as width is $\Delta r$ and length is $2\prod r$ .

  $\text{Area}=(2\prod r)\Delta r{\text{ in}}^2$

Conclusion:

The area of the strip is $(2\prod r)\Delta r{\text{ in}}^2$

(b)

To determine

The reason that oil passes through the ring at approximately $8(10-r^2)(2\prod r)\Delta r$ cubic inches per second.

Answer to Problem 34E

The reason that oil passes through the ring at approximately $8(10-r^2)(2\prod r)\Delta r$ cubic inches per second is because volume per second is equal to the flow in $\frac{i{n}^3}{\sec }$ .

Explanation of Solution

Given:

The radius of cylindrical pipe is $3\text{ inches}$ and the speed at which oil flows at a distance r from the center is $8(10-r^2)\text{ inches per}\mathrm{second}$ .

Calculation:

  $\begin{array}{l}\text{Volume per second = inches per second × cross section area}\\ {\text{ = 8(10 - r}}^2)\frac{in}{\sec }\times (2\prod r)\Delta r{\text{ in}}^2\\ {\text{ = (80 - 8r}}^2)(2\prod r\Delta r)\frac{i{n}^3}{\sec }\end{array}$

Therefore, volume per second is flow in $\frac{i{n}^3}{\sec }$ .

Conclusion:

The reason that oil passes through the ring at approximately $8(10-r^2)(2\prod r)\Delta r$ cubic inches per second is because volume per second is equal to the flow in $\frac{i{n}^3}{\sec }$ .

(c)

To determine

The rate at which oil is flowing through the pipe.

Answer to Problem 34E

The rate at which oil is flowing through the pipe is $\text{1243}\text{.44 }\frac{i{n}^3}{\sec }$ .

Explanation of Solution

Given:

The radius of cylindrical pipe is $3\text{ inches}$ and the speed at which oil flows at a distance r from the center is $8(10-r^2)\text{ inches per}\mathrm{second}$ .

Calculation:

  $\begin{array}{l}Rate={\displaystyle \underset{0}{\overset{3}{\int }}8(10-r^2)(2\prod r)dr}\\ \text{ = 16}\prod {\displaystyle \underset{0}{\overset{3}{\int }}(10r-r^3)}dr\\ \text{ =16}\prod {\left[10\left(\frac{r^2}{2}\right)-\left(\frac{r^4}{4}\right)\right]}_0^3\\ \text{ = 16}\prod \left[5{(3)}^2-\frac{1}{4}{(3)}^4\right]\\ \text{ = 16}\prod \left[5(9)-\frac{1}{4}(81)\right]\\ \text{ = 16}\prod \left[45-20.25\right]\\ \text{ = 16}\prod \times 24.75\\ \text{ = 1243}\text{.44 }\frac{i{n}^3}{\sec }\end{array}$

Conclusion:

The rate at which oil is flowing through the pipe is $\text{1243}\text{.44 }\frac{i{n}^3}{\sec }$ .