Chapter 6.2, Problem 53E

(a)

To determine

To Show: The radius of the barrel with the height and maximum radius $R$ is constructed by rotating about the x axis and the parabola of $y=R-c{x}^2,-\frac{h}{2}\le x\le \frac{h}{2}$ , here $c$ is the positive constant and the radius of each end of the barrel is $r=R-d$ where $d=\frac{c{h}^2}{4}$ .

Explanation of Solution

Calculation:

Consider the graph shown in Figure 1

  

Figure 1

Since, the function $y=R-c{x}^2$ is the even function so the graph is symmetric about the y axis so the radius of the both the ends are equal.

When the region is rotated bounded by the region bounded by the curve $y=R-c{x}^2$ from $x=\frac{-h}{2}$ to $x=\frac{h}{2}$ and about the x axis then the solid is obtained that is the barrel.

Then, consider the vertical strip in the region then after the rotation a circular disk is obtained that has the radius of $r=R-c{x}^2$ for $x=\frac{h}{2}$ that is,

  $\begin{array}{c}r\left(\frac{h}{2}\right)=R-c{\left(\frac{h}{2}\right)}^2\\ =R-\frac{c{h}^2}{4}\end{array}$

Consider $d=\frac{c{h}^2}{4}$ then $r=R-d$ .

Hence, proved.

(b)

To determine

To Show: The volume enclosed by the barrel is $V=\frac{1}{3}\pi h\left(2R^2+r^2-\frac{2}{5}{d}^2\right)$ .

Explanation of Solution

Calculation:

Consider the volume of the barrel is,

  $\begin{array}{c}V=2{\displaystyle {\int }_0^{\frac{h}{2}}\pi {\left(R-c{x}^2\right)}^{2}dx}\\ =2\pi {\displaystyle {\int }_0^{\frac{h}{2}}\left(R^2+c^2{x}^4-2Rc{x}^2\right)dx}\\ =2\pi \left[\frac{R^{2}h}{2}+\frac{1}{160}{c}^2{h}^5-\frac{2}{24}Rc{h}^3\right]\end{array}$

Then, multiply the numerator and the denominator with 3 as,

  $\begin{array}{c}V=\frac{\pi h}{3}\left[3R^2+\frac{3}{80}{c}^2{h}^4-\frac{1}{2}Rc{h}^2\right]\\ =\frac{1}{3}\pi h\left[2R^2+{\left[R-\left(\frac{c{h}^2}{4}\right)\right]}^2-\frac{1}{40}{c}^2{h}^4\right]\\ =\frac{1}{3}\pi h\left[2R^2+{\left[R-d\right]}^2-\frac{16}{40}\left(d^2\right)\right]\\ =\frac{1}{3}\pi h\left[2R^2+r^2-\frac{2}{5}{d}^2\right]\end{array}$

Hence, proved.