Q. 33 - Calculus

Question

Consider the region between the graph of $f\left(x\right)=\frac{3}{x}$ and the x-axis on [1,3]. For each line of rotation given in Exercises 31– 34, use definite integrals to find the volume of the resulting solid.

Step-by-Step Solution

Answer

The required volume is $12\mathrm{\pi }$

Step 1. Given Information

The given figure is     

Step 2. Calculation

Express the curve as inverse function,

$$\begin{gathered} y=\frac{3}{x} \\ x=\frac{3}{y} \\ g\left(y\right)=\frac{3}{y} \end{gathered}$$

For the x-interval of $[1,3]$, the corresponding interval of y-variable will be $[0,3]$

The region in the figure will form two types of washers when rotated about y-axis.

For the first washer in the y-interval of $[0,1]$, the external radius of each washer is $x=3$ and internal radius is 1.

The required volume for first washer is as follows,

$$\begin{gathered} V_1=\mathrm{\pi }{\int }_0^1\left(3^2-1\right)\mathrm{dy} \\ =\mathrm{\pi }{\int }_0^{1}8\mathrm{dy} \\ =8\mathrm{\pi }{\int }_0^1\mathrm{dy} \end{gathered}$$

Step 3. Calculation

For the second washer in the y-interval of $[1,3]$, the external radius of each washer is $g\left(y\right)$ and internal radius is 1.

The required volume for second washer is as follows,

$$\begin{gathered} V_2=\mathrm{\pi }{\int }_1^3\left({\left(\mathrm{g}\left(\mathrm{y}\right)\right)}^2-1^2\right)\mathrm{dy} \\ =\mathrm{\pi }{\int }_1^3\left({\left(\frac{3}{\mathrm{y}}\right)}^2-1^2\right)\mathrm{dy} \\ =\mathrm{\pi }{\int }_1^3\left(\frac{9}{{\mathrm{y}}^2}-1\right)\mathrm{dy} \\ =\mathrm{\pi }{\int }_1^3\left(9{\mathrm{y}}^{-2}-1\right)\mathrm{dy} \end{gathered}$$

Add the integrals to find the volume of solid revolution.

$$\begin{gathered} V=V_1+V_2 \\ =8\mathrm{\pi }{\int }_0^1\mathrm{dy}+\mathrm{\pi }{\int }_1^3(9{\mathrm{y}}^{-2}-1)\mathrm{dy} \\ =8\mathrm{\pi }{\left[\mathrm{y}\right]}_0^1+\mathrm{\pi }{\left[\frac{-9}{\mathrm{y}}-1\right]}_1^3 \\ =8\mathrm{\pi }+\mathrm{\pi }(-6+10) \\ =12\mathrm{\pi } \end{gathered}$$