Q. 64 - Calculus

Question

In Exercises 63–72, set up and solve a definite integral to find the exact area of each surface of revolution obtained by revolving the curve y = f(x) around the x-axis on the interval [a, b]. 

$f\left(x\right)=\frac{1}{x},[a,b]=[1,10]$

Step-by-Step Solution

Answer

The exact area of the surface of the revolution obtained by revolving the curve $f\left(x\right)=\frac{1}{x}$around the x-axis on the interval $\left[1,10\right]$ is $S=\pi \left[\sqrt{2}-\frac{\sqrt{1+(100{)}^2}}{100}-\ln (1+\sqrt{2})+\ln \left(1+\sqrt{1+(100{)}^2}\right)\right].$

Step 1. Given Information.

The given curve is $f\left(x\right)=\frac{1}{x}$ and the interval is $\left[1,10\right].$

Step 2. Find the exact area.

To find the area, we will use the formula of surface area as a definite integral which is $S=2\pi {\int }_a^{b}f\left(x\right)\sqrt{1+{(f\text{'}(x\left)\right)}^2}dx.$

So,

$$\begin{gathered} S=2\pi {\int }_1^{10}\frac{1}{x}\sqrt{1+{\left(-\frac{1}{x^2}\right)}^2}dx \\ S=2\pi {\int }_1^{10}\sqrt{1+\frac{1}{x^4}}\cdot x^2\cdot \frac{1}{x^3}dx \\ \mathrm{Now}, \mathrm{let} u=\frac{1}{x^2}, \left(-\frac{1}{2}\right)du=\left(\frac{1}{x^3}\right)dx \\ S=2\pi {\int }_1^{1/100}\sqrt{1+u^2}\cdot \frac{1}{u}\cdot \left(-\frac{1}{2}\right)du \\ S=-\pi {\int }_1^{1/100}\frac{\sqrt{1+u^2}}{u}du \\ S=-\pi {\left[\sqrt{1+u^2}-\ln \left|\frac{1+\sqrt{1+u^2}}{u}\right|\right]}_1^{1/100} \\ S=-\pi \left[\sqrt{1+{\left(\frac{1}{100}\right)}^2}-\ln \left|\frac{1+\sqrt{1+{\left(\frac{1}{100}\right)}^2}}{(1/100)}\right|-\sqrt{2}+\ln (1+\sqrt{2})\right] \\ S=\pi \left[\sqrt{2}-\frac{\sqrt{1+(100{)}^2}}{100}-\ln (1+\sqrt{2})+\ln \left(1+\sqrt{1+(100{)}^2}\right)\right] \end{gathered}$$

Thus, the exact area is $S=\pi \left[\sqrt{2}-\frac{\sqrt{1+(100{)}^2}}{100}-\ln (1+\sqrt{2})+\ln \left(1+\sqrt{1+(100{)}^2}\right)\right].$