Chapter 8.1, Problem 24E

a.

To determine

To state the approximate area.

Answer to Problem 24E

The area is $2\pi r\Delta \text{r}$

Explanation of Solution

Given information:

Radius of the ring is r and its thickness is $\Delta \text{r}$ .

If the ring is straighten out the circumference of the ring become the length of the rectangle and its thickness becomes breadth of the rectangle.

Therefore,

The area is $2\pi r\Delta \text{r}$

b.

To determine

To explain why we know the oil passes through the ring at approximately $8\left(10-r^2\right)\left(2\pi r\right)\Delta r$ cubic inches per second.

Answer to Problem 24E

  $\text{flow rate = speed of flow }\times \text{ area}$

Explanation of Solution

Given information:

Radius of the ring is r and its thickness is $\Delta \text{r}$ .

The flow rate of the ring is $8\left(10-r^2\right)\left(2\pi r\right)\Delta r$ because

  $\text{flow rate = speed of flow }\times \text{ area}$

And $8\left(10-r^2\right)$ represents speed of flow in inches per second, $\left(2\pi r\right)\Delta \text{r}$ represents area.

c.

To determine

To evaluate rate of flow of oil through the pipe using definite integral.

Answer to Problem 24E

The flow rate is approximately 1876.6 cubic inches per second.

Explanation of Solution

Given information:

Radius of the ring is 3 inches and its thickness is $\Delta \text{r}$ .

Calculation:

Seek the cumulative effect of speed of flow for $0\le r\le 3$

The flow rate is

  $\begin{array}{l}={\displaystyle \sum 8\left(10-r^2\right)2\pi r\Delta r}\text{ cubic inches per second}\\ ={\displaystyle \underset{0}{\overset{3}{\int }}8\left(10-r^2\right)2\pi r\Delta r}\text{ cubic inches per second}\\ =8{\left[20\pi r^2-\frac{2}{3}\pi r^3\right]}_0^2\text{ cubic inches per second}\\ =1876.57\text{ cubic inches per second}\end{array}$

Therefore,

The flow rate is approximately 1876.6 cubic inches per second.