find the volume of revolution shown on the right by rotating the area shown on the left about the x - axis…

Explanation:

Step1: Recall disk - method formula

The formula for the volume $V$ of the solid of revolution about the $x$ - axis using the disk method is $V=\pi\int_{a}^{b}[f(x)]^{2}dx$, where $y = f(x)$ is the function and $[a,b]$ is the interval of integration. Here, $f(x)=\frac{1}{x^{3}}$, $a = 1$, and $b = 2$.

Step2: Set up the integral

$V=\pi\int_{1}^{2}(\frac{1}{x^{3}})^{2}dx=\pi\int_{1}^{2}\frac{1}{x^{6}}dx$.

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 $n=-6$, we have $\int\frac{1}{x^{6}}dx=\int x^{-6}dx=\frac{x^{-6 + 1}}{-6+1}=-\frac{1}{5x^{5}}+C$.

Step4: Evaluate the definite integral

$V=\pi\left[-\frac{1}{5x^{5}}\right]_{1}^{2}=\pi\left(-\frac{1}{5\times2^{5}}+\frac{1}{5\times1^{5}}\right)$. $V=\pi\left(-\frac{1}{160}+\frac{1}{5}\right)=\pi\left(\frac{-1 + 32}{160}\right)=\frac{31\pi}{160}$.

Step5: Calculate the numerical value

$V=\frac{31\pi}{160}\approx\frac{31\times3.14159}{160}\approx0.609$.

Answer:

$0.609$