Find the volume of this improper integral?

Question

Find the volume of the described solid of revolution or state that it does not exist.

The region bounded by $f(x)=\sqrt{\frac{(x+1)}{x^3}}$ and the $x$-axis on the interval $[1,\infty)$ is resolved about the $x$-axis

I tried this many times but I keep getting $2\pi$.

Answer 1

Here are the steps $$ V=\pi\int_1^{\infty}\left(\sqrt{\frac{(x+1)}{x^3}}\right)^2 dx$$ $$ =\pi \lim\limits_{b\to\infty} \int_1^{b}\left|\frac{x+1}{x^3}\right| dx$$ $$ =\pi \lim\limits_{b\to\infty} \int_1^{b}\left(\frac{1}{x^2}+\frac{1}{x^3}\right) dx$$ $$ =\pi \lim\limits_{b\to\infty} \int_1^{b}x^{-2}dx+ \pi \lim\limits_{b\to\infty} \int_1^{b} x^{-3}dx$$ $$ =-\pi \lim\limits_{b\to\infty}x^{-1}\bigg|_1^{b}-\frac{\pi}{2} \lim\limits_{b\to\infty} x^{-2}\bigg|_1^{b} $$ $$ =-\pi \lim\limits_{b\to\infty}\frac{1}{x}\bigg|_1^{b}-\frac{\pi}{2} \lim\limits_{b\to\infty}\frac{1}{x^2}\bigg|_1^{b} $$ $$ =-\pi \lim\limits_{b\to\infty}\left[\frac{1}{b}- 1 \right]-\frac{\pi}{2} \lim\limits_{b\to\infty}\left[\frac{1}{b^2}-1\right]$$ $$ =-\pi (0- 1)-\frac{\pi}{2}(0-1)$$ $$ =-\pi (- 1)-\frac{\pi}{2}(-1)$$ $$ =\pi +\frac{\pi}{2}=\frac{3\pi}{2} $$

Answer 2

The volume of revolution is formed by rotating $f(x)=\sqrt{(x+1)/x^3}$ around the $x$-axis along $[1,\infty)$. As such, we can proceed with the standard formula,

$$V=\pi \int_1^\infty \! f(x)^2 \, dx = \pi \int_1^\infty \left( \frac{1}{x^2} + \frac{1}{x^3}\right) \, dx =\pi \left( \frac{3}{2} - \lim_{x\to \infty} \frac{1+2x}{2x^2}\right) = \frac{3\pi}{2}.$$