Chapter 6.2, Problem 29E

a)

To determine

To calculate: The volume of the solid region rotated about the x-axis using the midpoint rule.

Answer to Problem 29E

The volume of the solid region rotated about the x-axis is 195.66.

Explanation of Solution

Given information:

Number of intervals ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}\approx {\displaystyle \sum _{i=1}^{n}f\left({\overline{x}}_i\right)\Delta x}=\Delta x\left(f\left({\overline{x}}_1\right)+\mathrm{.......}+f\left({\overline{x}}_n\right)\right)$.

The region lies between $\Delta x=\frac{b-a}{n}$ and ${\overline{x}}_i=\frac{1}{2}\left(x_{i-1}+x_i\right)$.

The endpoints are, 2, 4, 6, 8, and 10.

The midpoints are, 3, 5, 7, and 9.

Apply midpoint rule:

$\Delta x$

Where, $\left(x_{i-1},x_i\right)$ and ${\overline{x}}_i$

Here, width of the subintervals is $n=4$, the upper limit is b, the lower limit is a and the midpoint of $a=2$ is $b=10$.

Calculation:

The expression to find the volume as shown below.

$V={\displaystyle \underset{a}{\overset{b}{\int }}A\left(x\right)dx}$ (1)

The expression to find the volume of the solid region using midpoint rule as shown below.

$\begin{array}{c}{\displaystyle \underset{a}{\overset{b}{\int }}A\left(x\right)dx}=\frac{b-a}{n}\left(A\left(3\right)+A\left(5\right)+A\left(7\right)+A\left(9\right)\right)\\ =\frac{b-a}{n}\left(\pi {\left(f\left(3\right)\right)}^2+\pi {\left(f\left(5\right)\right)}^2+\pi {\left(f\left(7\right)\right)}^2+\pi {\left(f\left(9\right)\right)}^2\right)\end{array}$ (2)

Substitute 2 for a, 10 m for b, 4 for n, 1.5 for $f\left(3\right)$, 2.2 for $f\left(5\right)$, 3.8 for $f\left(7\right)$, and 3.1 for $f\left(9\right)$ in Equation (2).

$\begin{array}{c}{\displaystyle \underset{a}{\overset{b}{\int }}A\left(x\right)dx}=\frac{10-2}{4}\left(\left(\pi \times {1.5}^2\right)+\left(\pi \times {2.2}^2\right)+\left(\pi \times {3.8}^2\right)+\left(\pi \times {3.1}^2\right)\right)\\ =2\pi \times \left(2.25+4.84+14.44+9.61\right)\\ =2\pi \times 31.91\\ =195.66\end{array}$

Therefore, the volume of the solid region rotated about the x-axis is 195.66.

b)

To determine

To calculate: The volume of the solid region rotated about the y-axis using the midpoint rule.

Answer to Problem 29E

The volume of the solid region rotated about the y-axis is 838.27.

Explanation of Solution

Given information:

Number of intervals $n=4$.

The region lies between $a=0$ and $b=4$.

The endpoints are, 0, 1, 2, 3 and 4.

The midpoints are, 0.5, 1.5, 2.5, and 3.5.

Calculation:

The expression to find the volume of the solid region using midpoint rule as shown below.

$\begin{array}{c}{\displaystyle \underset{a}{\overset{b}{\int }}A\left(x\right)dx}=\frac{b-a}{n}\left(A\left(0.5\right)+A\left(1.5\right)+A\left(2.5\right)+A\left(3.5\right)\right)\\ =\frac{b-a}{n}\left(\pi {\left(r_{out}^2-r_{in}^2\right)}_{0.5}+\pi {\left(r_{out}^2-r_{in}^2\right)}_{1.5}+\pi {\left(r_{out}^2-r_{in}^2\right)}_{2.5}+\pi {\left(r_{out}^2-r_{in}^2\right)}_{3.5}\right)\end{array}$ (3)

Sketch the region as shown in Figure 1.

Refer to Figure 1.

Show outer radius and inner radius of the midpoints as follows:

At the midpoint 0.5.

$\begin{array}{l}{r}_{out}=9.9\\ r_{in}=2.2\end{array}$

At the midpoint 1.5.

$\begin{array}{l}{r}_{out}=9.7\\ r_{in}=3.0\end{array}$

At the midpoint 2.5.

$\begin{array}{l}{r}_{out}=9.3\\ r_{in}=5.6\end{array}$

At the midpoint 3.5.

$\begin{array}{l}{r}_{out}=8.7\\ r_{in}=6.5\end{array}$

Substitute 0 for a, 4 for b, 4 for n, and substitute the known values in Equation (4).

$\begin{array}{c}{\displaystyle \underset{a}{\overset{b}{\int }}A\left(x\right)dx}=\frac{4-0}{4}\times \pi \left[\left({9.9}^2-{2.2}^2\right)+\left({9.7}^2-{3.0}^2\right)+\left({9.3}^2-{5.6}^2\right)+\left({8.7}^2-{6.5}^2\right)\right]\\ =\pi \times \left(93.17+85.09+55.13+33.44\right)\\ =\pi \times 266.83\\ =838.27\end{array}$

Therefore, the volume of the solid region rotated about the y-axis is 838.27.