91影视

The given figure is

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 disks when rotated about y-axis.

For the first disk in the y-interval of $[0,1]$, the radius of each disk is 2.

The required volume for first disk is as follows,

$$\begin{gathered} V_1=\mathrm{蟺}{鈭�}_0^1\left(2^2\right)\mathrm{dy} \\ =4\mathrm{蟺}{鈭�}_0^1\mathrm{dy} \end{gathered}$$

For the second disk in the y-interval of $[1,3]$, the radius of each disk is $g\left(y\right)-1$

The required volume for second disk is as follows,

$$\begin{gathered} V_2=\mathrm{蟺}{鈭�}_1^3{\left(\mathrm{g}\left(\mathrm{y}\right)-1\right)}^2\mathrm{dy} \\ =\mathrm{蟺}{鈭�}_1^3{\left(\frac{3}{\mathrm{y}}-1\right)}^2\mathrm{dy} \\ =\mathrm{蟺}{鈭�}_1^3\left(\frac{9}{{\mathrm{y}}^2}-\frac{6}{\mathrm{y}}+1\right)\mathrm{dy} \\ =\mathrm{蟺}{鈭�}_1^3\left(9{\mathrm{y}}^{-2}-\frac{6}{\mathrm{y}}+1\right)\mathrm{dy} \end{gathered}$$

Add the integrals to find the volume of solid revolution.

$$\begin{gathered} V=V_1+V_2 \\ =4\mathrm{蟺}{鈭�}_0^1\mathrm{dy}+\mathrm{蟺}{鈭�}_1^3\left(9{\mathrm{y}}^{-2}-\frac{6}{\mathrm{y}}+1\right)\mathrm{dy} \\ =4\mathrm{蟺}{\left[\mathrm{y}\right]}_0^1+\mathrm{蟺}{\left[-\frac{9}{\mathrm{y}}-6\mathrm{lny}+\mathrm{y}\right]}_1^3 \\ =4\mathrm{蟺}(1-0)+\mathrm{蟺}\left(\left(-\frac{9}{3}-6\ln 3+3\right)-\left(-\frac{9}{1}-6\ln 1+1\right)\right) \\ =4\mathrm{蟺}+\mathrm{蟺}\left(\left(-3-6\ln 3+3\right)-\left(-9-6\left(0\right)+1\right)\right) \\ =4\mathrm{蟺}-6\mathrm{蟺濒苍}3+8\mathrm{蟺} \\ =\mathrm{蟺}(12-6\ln 3) \end{gathered}$$