Question

The area enclosed by the graphs of y = 1/x, y = 1, and x = 3 is rotated about the line y = -1. Find the volume and show steps.

Answer 1

Answer:

$A=\pi\displaystyle\biggr[\frac{16}{3}-2\ln(|3|)\biggr]\approx9.8524$

Step-by-step explanation:

Use the Washer Method $A=\pi\displaystyle \int\limits_{a}^{b}{\bigr[R(x)^2-r(x)^2\bigr] \, dx$ where $R(x)$ is the outer radius and $r(x)$ is the inner radius.

If we sketch out the graph, we see that $y=1$ intersects points $(1,1)$ and $(3,1)$, which will be our bounds of integration.

Here, our outer radius will be $R(x)=(-1-1)=-2$ and our inner radius will be $r(x)=-1-\frac{1}{x}$.

Thus, we can compute the integral and find the volume:

$A=\pi\displaystyle\int\limits^{3}_{1} {(-2)^2-\biggr(-1-\frac{1}{x}\biggr)^2 } \, dx\\ \\A=\pi\displaystyle\int\limits^{3}_{1} {4-\biggr(1+\frac{2}{x}+\frac{1}{x^2} \biggr) } \, dx\\\\A=\pi\displaystyle\int\limits^{3}_{1} {4-1-\frac{2}{x}-\frac{1}{x^2}} \, dx\\\\A=\pi\displaystyle\int\limits^{3}_{1} {3-\frac{2}{x}-\frac{1}{x^2}} \, dx\\\\A=\pi\displaystyle\biggr[3x-2\ln(|x|)+\frac{1}{x}\biggr]\Biggr|_{1}^{3}$

$A=\pi\displaystyle\biggr[\biggr(3(3)-2\ln(|3|)+\frac{1}{3}\biggr)-\biggr(3(1)-2\ln(|1|)+\frac{1}{1}\biggr)\biggr]\\\\A=\pi\displaystyle\biggr[\biggr(9-2\ln(|3|)+\frac{1}{3}\biggr)-\biggr(3+1\biggr)\biggr]\\\\A=\pi\displaystyle\biggr[\biggr(\frac{28}{3}-2\ln(|3|)\biggr)-\biggr(4\biggr)\biggr]\\A=\pi\displaystyle\biggr[\frac{16}{3}-2\ln(|3|)\biggr]\\A\approx9.8524$

In conclusion, the volume of the solid of revolution will be about 9.8524 cubic units. See the attached graph for a helpful visual!