find the volume of the described solid of revolution or state that it does not exist. the region bounded by…

Explanation:

Step1: Recall volume - of - revolution formula

The volume $V$ of the solid of revolution about the $x$ - axis using the disk method for a function $y = f(x)$ from $x=a$ to $x = b$ is given by $V=\pi\int_{a}^{b}[f(x)]^{2}dx$. Here, $a = 2$, $b=\infty$, and $f(x)=x^{-5}$. So, $V=\pi\int_{2}^{\infty}(x^{-5})^{2}dx=\pi\int_{2}^{\infty}x^{- 10}dx$.

Step2: Evaluate the improper integral

The improper integral $\int_{2}^{\infty}x^{-10}dx=\lim_{t\rightarrow\infty}\int_{2}^{t}x^{-10}dx$. Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $\lim_{t\rightarrow\infty}\int_{2}^{t}x^{-10}dx=\lim_{t\rightarrow\infty}\left[\frac{x^{-10 + 1}}{-10+1}\right]{2}^{t}=\lim{t\rightarrow\infty}\left[-\frac{1}{9x^{9}}\right]_{2}^{t}$.

Step3: Compute the limit

$\lim_{t\rightarrow\infty}\left(-\frac{1}{9t^{9}}+\frac{1}{9\times2^{9}}\right)$. As $t\rightarrow\infty$, $\frac{1}{9t^{9}}\rightarrow0$. So, $\lim_{t\rightarrow\infty}\left(-\frac{1}{9t^{9}}+\frac{1}{9\times2^{9}}\right)=\frac{1}{9\times512}=\frac{1}{4608}$.

Step4: Find the volume

Since $V=\pi\int_{2}^{\infty}x^{-10}dx$, then $V=\frac{\pi}{4608}$.

Answer:

A. The volume is $\frac{\pi}{4608}$ cubic units.