Chapter 6.3, Problem 36E

To determine

To find: The volume of the solid (torus of Exercise 6.2.63) using cylindrical shell.

Answer to Problem 36E

Thevolume of the solid (torus) is $\underset{\_}{2{\pi }^{2}R{r}^2}$.

Explanation of Solution

Given:

Solid torus with radii r and R.

The equation of the circle is ${\left(x-R\right)}^2+y^2=r^2$

Show the theorem 7 (Integrals of symmetric Functions):

Apply the condition as follows if the function f is continuous on [–a, a].

1. a) If f is even $\left[f\left(-x\right)=f\left(x\right)\right]$, then ${\displaystyle {\int }_{-a}^{a}f\left(x\right)dx}=2{\displaystyle {\int }_0^{a}f\left(x\right)dx}$.
2. b) If f is odd $\left[f\left(-x\right)=-f\left(x\right)\right]$, then ${\displaystyle {\int }_{-a}^{a}f\left(x\right)dx}=0$.

Calculation:

Show the equation as below:

$\begin{array}{c}{\left(x-R\right)}^2+y^2=r^2\\ y^2=r^2-{\left(x-R\right)}^2\\ y=\sqrt{r^2-{\left(x-R\right)}^2}\end{array}$ (1)

The region lies between $x=R+r$ and $x=R-r$.

Sketch the solid region as shown in Figure 1.

Calculate the volume using the method of cylindricalshell:

$V={\displaystyle {\int }_a^{b}2\pi x\left[f\left(x\right)\right]dx}$ (2)

Substitute $R+r$ for a,$R-r$ for b, and $2\sqrt{r^2-{\left(x-R\right)}^2}$ for $\left[f\left(x\right)\right]$ in Equation (2).

$\begin{array}{c}V={\displaystyle {\int }_{R-r}^{R+r}2\pi x2\sqrt{r^2-{\left(x-R\right)}^2}dx}\\ =4\pi {\displaystyle {\int }_{R-r}^{R+r}x\sqrt{r^2-{\left(x-R\right)}^2}dx}\end{array}$ (3)

Consider $u=x-R$ (4)

Differentiate both sides of the equation.

$du=dx$

Rearrange Equation (4).

$\begin{array}{c}u+R=x\\ x=u+R\end{array}$

Calculate the lower limit value of u using Equation (4).

Substitute $R+r$ for x in Equation (4).

$\begin{array}{c}u=R+r-R\\ =r\end{array}$

Calculate the upper limit value of u using Equation (4).

Substitute $R-r$ for x in Equation (4).

$\begin{array}{c}u=R-r-R\\ =-r\end{array}$

Apply lower and upper limits for u in Equation (3).

Substitute $u-R$ for x, u for $\left(r-R\right)$, and du for dx in Equation (3).

$\begin{array}{c}V=4\pi {\displaystyle {\int }_{-r}^r\left(u+R\right)\sqrt{r^2-u^2}du}\\ =4\pi {\displaystyle {\int }_{-r}^r\left(u\sqrt{r^2-u^2}+R\sqrt{r^2-u^2}\right)du}\\ =4\pi {\displaystyle {\int }_{-r}^{r}u\sqrt{r^2-u^2}du}+4\pi {\displaystyle {\int }_{-r}^{r}R\sqrt{r^2-u^2}du}\end{array}$ (5)

Consider $V_1$ and $V_2$ as follows:

$V_1=4\pi {\displaystyle {\int }_{-r}^{r}u\sqrt{r^2-u^2}du}$ (6)

$V_2=4\pi {\displaystyle {\int }_{-r}^{r}R\sqrt{r^2-u^2}du}$ (7)

Substitute $V_1$ for $4\pi {\displaystyle {\int }_{-r}^{r}u\sqrt{r^2-u^2}du}$ and $V_2$ for $4\pi {\displaystyle {\int }_{-r}^{r}R\sqrt{r^2-u^2}du}$ in Equation (5).

$V=V_1+V_2$ (8)

Consider $f\left(u\right)=u\sqrt{r^2-u^2}$ (9)

Substitute $\left(-u\right)$ for u in Equation (9).

$\begin{array}{c}f\left(u\right)=-u\sqrt{r^2-{\left(-u\right)}^2}\\ =-u\sqrt{r^2-u^2}\\ =-f\left(u\right)\end{array}$

Apply the theorem in Equation (6).

The function satisfies the condition as $\left[f\left(-u\right)=-f\left(u\right)\right]$.

Thus, the function is an odd function.

Therefore, the volume $V_1$ is 0.

Integrate Equation (7).

$\begin{array}{c}V=4\pi R{\left[\frac{1}{2}r\sqrt{r^2}\left(\arcsin \left(\frac{u}{r}\right)+\frac{1}{2}\sin \left(2\arcsin \left(\frac{u}{r}\right)\right)\right)\right]}_{-r}^r\\ =\left\{4\pi R\left\{\begin{array}{l}\left[\frac{1}{2}{r}^2\left(\arcsin \left(\frac{r}{r}\right)+\frac{1}{2}\sin \left(2\arcsin \left(\frac{r}{r}\right)\right)\right)\right]-\\ \left[\frac{1}{2}{r}^2\left(\arcsin \left(\frac{-r}{r}\right)+\frac{1}{2}\sin \left(2\arcsin \left(\frac{-r}{r}\right)\right)\right)\right]\end{array}\right\}\right\}\\ =4\pi R\left\{\left[\frac{1}{2}{r}^2\left(\frac{\pi }{2}+\frac{1}{2}\sin \left(\frac{2\pi }{2}\right)\right)\right]-\left[\frac{1}{2}{r}^2\left(-\frac{\pi }{2}+\frac{1}{2}\sin \left(\frac{-2\pi }{2}\right)\right)\right]\right\}\end{array}$

$\begin{array}{c}=4\pi R\left(\frac{\pi }{4}{r}^2+0+\frac{\pi }{4}{r}^2+0\right)\\ =4\pi R\left(\frac{\pi }{2}{r}^2\right)\\ =2{\pi }^{2}R{r}^2\end{array}$

The volume is,

Substitute 0 for $A_1$ and $2{\pi }^{2}R{r}^2$ for $A_2$ in Equation (8)

$\begin{array}{c}V=0+2{\pi }^{2}R{r}^2\\ =2{\pi }^{2}R{r}^2\end{array}$

Hence, the volume of the solid (torus) is $\underset{\_}{2{\pi }^{2}R{r}^2}$.