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 is given by $V=\pi\int_{a}^{b}[f(x)]^{2}dx$. Here, $a = 1$, $b=\infty$, and $f(x)=x^{-7}$. So, $V=\pi\int_{1}^{\infty}(x^{-7})^{2}dx=\pi\int_{1}^{\infty}x^{- 14}dx$.

Step2: Evaluate the improper integral

The improper integral $\int_{1}^{\infty}x^{-14}dx=\lim_{t\rightarrow\infty}\int_{1}^{t}x^{-14}dx$. Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $\lim_{t\rightarrow\infty}\int_{1}^{t}x^{-14}dx=\lim_{t\rightarrow\infty}\left[\frac{x^{-14 + 1}}{-14+1}\right]{1}^{t}=\lim{t\rightarrow\infty}\left[\frac{x^{-13}}{-13}\right]_{1}^{t}$.

Step3: Calculate the limit

$\lim_{t\rightarrow\infty}\left[\frac{x^{-13}}{-13}\right]{1}^{t}=\lim{t\rightarrow\infty}\left(-\frac{1}{13t^{13}}+\frac{1}{13}\right)$. As $t\rightarrow\infty$, $\frac{1}{13t^{13}}\rightarrow0$. So, $\lim_{t\rightarrow\infty}\left(-\frac{1}{13t^{13}}+\frac{1}{13}\right)=\frac{1}{13}$.

Step4: Find the volume

Since $V=\pi\int_{1}^{\infty}x^{-14}dx$, and $\int_{1}^{\infty}x^{-14}dx=\frac{1}{13}$, then $V = \frac{\pi}{13}$.

Answer:

A. The volume is $\frac{\pi}{13}$ cubic units.