find the volume of the solid generated by revolving the region bounded by the given lines and curves about…

Explanation:

Step1: Recall volume - of - revolution formula

The formula for the volume $V$ of the solid generated by revolving the region bounded by $y = f(x)$, $y = 0$, $x=a$, and $x = b$ about the $x$-axis is $V=\pi\int_{a}^{b}[f(x)]^{2}dx$. Here, $f(x)=\frac{1}{x}$, $a = 1$, and $b = 2$.

Step2: Set up the integral

Substitute $f(x)=\frac{1}{x}$, $a = 1$, and $b = 2$ into the formula: $V=\pi\int_{1}^{2}(\frac{1}{x})^{2}dx=\pi\int_{1}^{2}\frac{1}{x^{2}}dx=\pi\int_{1}^{2}x^{-2}dx$.

Step3: Integrate the function

Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), for $n=-2$, we have $\int x^{-2}dx=-\frac{1}{x}+C$. Then $\pi\int_{1}^{2}x^{-2}dx=\pi\left[-\frac{1}{x}\right]_{1}^{2}$.

Step4: Evaluate the definite integral

$\pi\left[-\frac{1}{x}\right]_{1}^{2}=\pi\left(-\frac{1}{2}+\frac{1}{1}\right)=\pi\left(1 - \frac{1}{2}\right)=\frac{1}{2}\pi$.

Answer:

B. $\frac{1}{2}\pi$