Chapter 6.2, Problem 54E

To determine

Find the volume of the curve

Answer to Problem 54E

The volume of the curve is $2k\pi A$

Explanation of Solution

Given information:

Calculation:

Let $f\left(x\right)$ be the upper border of the region and $g\left(x\right)$ be the lower border.

In other words $f\left(x\right)>g\left(x\right)$ for interval $\left[a,b\right]$ .

  

Given region $R$ has area $A$

  $A={\displaystyle \underset{a}{\overset{b}{\int }}\left(f\left(x\right)-g\left(x\right)\right)}dx$

Region $R$ rotated about the $x-$ axis has volume $V_1$ (with disk method):

  $V_1=\pi {\displaystyle \underset{a}{\overset{b}{\int }}\left({\left[f\left(x\right)\right]}^2-{\left[g\left(x\right)\right]}^2\right)}dx$

If region $R$ is rotated about the $y=-k$ instead then using the washer method:

The first integral is the same as $V_1$ and the second is $A$ ,so can substitute

  $V_2=V_1+2k\pi A$

  $\begin{array}{l}{V}_2=\pi {\displaystyle \underset{a}{\overset{b}{\int }}\left({\left[f\left(x\right)-\left(-k\right)\right]}^2-{\left[g\left(x\right)-\left(-k\right)\right]}^2\right)}dx\\ =\pi {\displaystyle \underset{a}{\overset{b}{\int }}\left({\left[f\left(x\right)+k\right]}^2-{\left[g\left(x\right)+k\right]}^2\right)}dx\\ =\pi {\displaystyle \underset{a}{\overset{b}{\int }}\left({\left[f\left(x\right)\right]}^2+2kf\left(x\right)+k^2-\left[g{\left(x\right)}^2+2kg\left(x\right)+k^2\right]\right)}dx\\ =\pi {\displaystyle \underset{a}{\overset{b}{\int }}\left({\left[f\left(x\right)\right]}^2+2kf\left(x\right)+k^2-g{\left(x\right)}^2-2kg\left(x\right)-k^2\right)}dx\\ =\pi {\displaystyle \underset{a}{\overset{b}{\int }}\left({\left[f\left(x\right)\right]}^2+2kf\left(x\right)-g{\left(x\right)}^2-2kg\left(x\right)\right)}dx\\ =\pi {\displaystyle \underset{a}{\overset{b}{\int }}\left({\left[f\left(x\right)\right]}^2-g{\left(x\right)}^{2}dx+\pi {\displaystyle \underset{a}{\overset{b}{\int }}2kf\left(x\right)-2kg\left(x\right)}dx\right)}\\ =\pi {\displaystyle \underset{a}{\overset{b}{\int }}\left({\left[f\left(x\right)\right]}^2-g{\left(x\right)}^{2}dx+2k\pi {\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)-g\left(x\right)}dx\right)}\end{array}$

The first integral is the same as $V_1$ and the second is $A$ ,so can substitute

  $V_2=V_1+2k\pi A$

Thus, the volume of the curve is $2k\pi A$