Q. 60 - Calculus

Question

In Exercises 57–62, use n frustums to approximate the area of the surface of revolution obtained by revolving the curve y = f(x) around the x-axis on the interval [a, b].

$f\left(x\right)=\ln x,[a,b]=[1,4]n=3$

Step-by-Step Solution

Answer

The area of the surface of revolution obtained by revolving the given curve on the given interval is $\pi \left[\begin{array}{c}\ln 2\cdot \sqrt{1+(\ln 2{)}^2}+(\ln 2+\ln 3)\cdot \sqrt{1+(\ln 3-\ln 2{)}^2}\\ +(\ln 3+\ln 4)\cdot \sqrt{1+(\ln 4-\ln 3{)}^2}\end{array}\right].$

Step 1. Given Information.

The given curve is $f\left(x\right)=\ln x$ and the interval is $\left[1,4\right].$

Step 2. Find the area of the surface.

We have to use 3 frustums to approximate the area of the surface of the revolution obtained by revolving the given curve on the given interval. To approximate the area, we will use the formula of the area of a surface of revolution which is $S=\sum _{k=1}^{n}2\pi r_k{s}_k.$

Here,

 $$\begin{gathered} s_k=\sqrt{{\left(\mathrm{\Delta x}\right)}^2+{\left({\mathrm{\Delta y}}_{\mathrm{k}}\right)}^2} , r_k=\frac{f\left(x_{k-1}\right)+f\left(x_k\right)}{2}, \Delta x=\frac{b-a}{n}, and \\ \mathrm{\Delta }{y}_k=f\left(x_k\right)-f\left(x_{k-1}\right) \text{for }k=0,1,2,\dots ,n. \end{gathered}$$

So,

 $$\begin{gathered} \mathrm{\Delta }x=\frac{b-a}{n} \\ \mathrm{\Delta }x=\frac{4-1}{3} \\ \mathrm{\Delta }x=1 \end{gathered}$$

Now, we have to find $x_k \mathrm{and} f\left(x_k\right) \mathrm{for} k=0,1,2,3:$

$\begin{array}{rl}{x}_0& =1,x_1=2,x_2=3,x_3=4\\ f\left(1\right)& =0,f\left(2\right)=\ln 2,f\left(3\right)=\ln 3,f\left(4\right)=\ln 4\end{array}$

Step 3. Solve.

Let's find $r_k and s_k:$

$$\begin{gathered} r_1=\frac{f\left(1\right)+f\left(2\right)}{2} \mathrm{and} s_1=\sqrt{(1{)}^2+(\ln 2-0{)}^2} \\ {r}_1=\frac{1}{2}\ln 2 s_1=\sqrt{1+{\left(\ln 2\right)}^2} \end{gathered}$$

And

$$\begin{gathered} r_2=\frac{f\left(2\right)+f\left(3\right)}{2} \mathrm{and} s_2=\sqrt{{\left(1\right)}^2+{\left(\ln 3-\ln 2\right)}^2} \\ {r}_2==\frac{\ln 2+\ln 3}{2} s_2=\sqrt{1+{\left(\ln 3-\ln 2\right)}^2} \\ \mathrm{And} \\ {r}_3=\frac{f\left(3\right)+f\left(4\right)}{2} \mathrm{and} s_3=\sqrt{{\left(1\right)}^2+{\left(\ln 4-\ln 3\right)}^2} \\ {r}_3=\frac{\ln 3+\ln 4}{2} s_3=\sqrt{1+{\left(\ln 4-\ln 3\right)}^2} \end{gathered}$$

Now, let's find the area of the surface:

$\begin{array}{rl}S& =2\pi r_1{s}_1+2\pi r_2{s}_2+2\pi r_3{s}_3\\ S& =2\pi \left[\begin{array}{c}\frac{1}{2}\ln 2\cdot \sqrt{1+(\ln 2{)}^2}+\frac{1}{2}(\ln 2+\ln 3)\cdot \sqrt{1+(\ln 3-\ln 2{)}^2}\\ +\frac{1}{2}(\ln 3+\ln 4)\cdot \sqrt{1+(\ln 4-\ln 3{)}^2}\end{array}\right]\\ S& =\pi \left[\begin{array}{c}\ln 2\cdot \sqrt{1+(\ln 2{)}^2}+(\ln 2+\ln 3)\cdot \sqrt{1+(\ln 3-\ln 2{)}^2}\\ +(\ln 3+\ln 4)\cdot \sqrt{1+(\ln 4-\ln 3{)}^2}\end{array}\right]\end{array}$