Solutions - 0130-03

Volume:

$$\begin{array}{c}{V}_G={\int }_1^{\infty }\pi {\left(\frac{1}{x}\right)}^{2}dx\\ \\ \gg \gg \underset{R\to \infty }{\lim }{\int }_1^R\pi {\left(\frac{1}{x}\right)}^{2}dx\\ \\ \gg \gg \underset{R\to \infty }{\lim }\pi (-\frac{1}{R}+1)\gg \gg \pi \end{array}$$

Surface area:

$$S_G={\int }_1^{\infty }2\pi \left(\frac{1}{x}\right)\sqrt{1+{(-1/x^2)}^2}dx$$

But notice this:

$$\left(\frac{1}{x}\right)\sqrt{1+{(-1/x^2)}^2}>\frac{1}{x}$$

But ${\displaystyle {\int }_1^{\infty }2\pi \frac{1}{x}dx}$ diverges!

So by the comparison test, $S_G$ diverges as well.