For an assignment we have been asked to compute the surface area of Torricelli's trumpet which is obtained by revolving $y=1/x$ where $x>=1$ about the x axis. We have to calculate the surface area from $x=1$ to $x=a$ where $a$ is a real number bigger than one. I cannot manage to work it out myself so I have had a search around and found a few examples of how it is calculated from $x=1$ to $x=\infty$. Here is the working to one:
SA = \int_1^\infty 2\pi y \sqrt{1+(y')^2}\ dx >2\pi\int_1^\infty ydx $=2\pi \int_1^\infty dx/x $ $=2\pi [\ln x]_1^\infty$ $=2\pi [\ln \infty-0]$ $=\infty$
So I understand that the formula for the area of the surface formed by revolving about the $x$-axis a smooth curve $y=f(x)$ from $x=1$ to $x=\infty$ is \int_1^\infty 2\pi y \sqrt{1+(y')^2}\ dx but I cannot for the life of me figure out why they have put that $>2\pi\int ydx$ and why they have consequently ignored the formula for surface area on the next line and have only integrated $\int 2\pi y dx $