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Answer:
[tex]A=\pi\displaystyle\biggr[\frac{16}{3}-2\ln(|3|)\biggr]\approx9.8524[/tex]
Step-by-step explanation:
Use the Washer Method [tex]A=\pi\displaystyle \int\limits_{a}^{b}{\bigr[R(x)^2-r(x)^2\bigr] \, dx[/tex] where [tex]R(x)[/tex] is the outer radius and [tex]r(x)[/tex] is the inner radius.
If we sketch out the graph, we see that [tex]y=1[/tex] intersects points [tex](1,1)[/tex] and [tex](3,1)[/tex], which will be our bounds of integration.
Here, our outer radius will be [tex]R(x)=(-1-1)=-2[/tex] and our inner radius will be [tex]r(x)=-1-\frac{1}{x}[/tex].
Thus, we can compute the integral and find the volume:
[tex]A=\pi\displaystyle\int\limits^{3}_{1} {(-2)^2-\biggr(-1-\frac{1}{x}\biggr)^2 } \, dx\\ \\A=\pi\displaystyle\int\limits^{3}_{1} {4-\biggr(1+\frac{2}{x}+\frac{1}{x^2} \biggr) } \, dx\\\\A=\pi\displaystyle\int\limits^{3}_{1} {4-1-\frac{2}{x}-\frac{1}{x^2}} \, dx\\\\A=\pi\displaystyle\int\limits^{3}_{1} {3-\frac{2}{x}-\frac{1}{x^2}} \, dx\\\\A=\pi\displaystyle\biggr[3x-2\ln(|x|)+\frac{1}{x}\biggr]\Biggr|_{1}^{3}\\[/tex]
[tex]A=\pi\displaystyle\biggr[\biggr(3(3)-2\ln(|3|)+\frac{1}{3}\biggr)-\biggr(3(1)-2\ln(|1|)+\frac{1}{1}\biggr)\biggr]\\\\A=\pi\displaystyle\biggr[\biggr(9-2\ln(|3|)+\frac{1}{3}\biggr)-\biggr(3+1\biggr)\biggr]\\\\A=\pi\displaystyle\biggr[\biggr(\frac{28}{3}-2\ln(|3|)\biggr)-\biggr(4\biggr)\biggr]\\A=\pi\displaystyle\biggr[\frac{16}{3}-2\ln(|3|)\biggr]\\A\approx9.8524[/tex]
In conclusion, the volume of the solid of revolution will be about 9.8524 cubic units. See the attached graph for a helpful visual!
