Torricelli's/Gabriel's Trumpet Surface Area 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 $$\int 2\pi y dx $$
 A: We will find 
$$\int \frac{2}{x}\sqrt{1+\frac{1}{x^4}}\; dx.$$
In what follows, we assume that $x>0$.  This is fine for your problem. And the $\pi$ is missing, because it is unpleasant to carry around.  
There are sneaky tricks that we could use, particularly after finding the answer.  But instead we will do what comes naturally, to show that the integral is not all that difficult.
A little manipulation shows that our integral is equal to 
$$\int \frac{2x}{x^4}\sqrt{1+x^4}\; dx.$$ 
Let $x^2=\tan\theta$.  Then $2x\,dx=\sec^2\theta\,d\theta\;$ and $\sqrt{1+x^4}=\sec\theta$.   We arrive at
$$\int \frac{\sec^3\theta}{\tan^2\theta}\,d\theta=\int\frac{1}{\sin^2\theta\cos\theta}\;d\theta=\int\frac{\cos\theta}{\sin^2\theta(1-\sin^2\theta)}\;d\theta.$$
It is all downhill from now on. Let $t=\sin\theta$.  We arrive at
$$\int \frac{1}{t^2(1-t^2)}\;dt=\int \left(\frac{1}{1-t^2}+\frac{1}{t^2}\right)\;dt.$$
The final result is 
$$\frac{1}{2}\ln(1+t) +\frac{1}{2}\ln(1-t)-\frac{1}{t} +C.$$
Unwind the substitutions.  If we feel like it, we can make considerable simplifications along the way.  
A fancier way:  Hyperbolic function substitutions used to be taught in first year calculus, but now seldom are. Such a substitution works very nicely here. 
Define functions $\cosh t$, $\sinh t$ by
$$\cosh t=\frac{e^t+e^{-t}}{2} \qquad\text{and}\qquad \sinh t =\frac{e^t-e^{-t}}{2}.$$
It is easy to see that the derivative of $\cosh t$ is $\sinh t$, and the derivative of $\sinh t$ is $\cosh t$.  Also, we have the useful identity $1+\sinh^2 t=\cosh^2 t$.  The other hyperbolic functions are defined in analogy with the way the other trigonometric functions are defined in terms of $\cos$ and $\sin$.  The remaining facts that we need below are easy to verify.
Let $x^2=\sinh t$.  Pretty quickly we end up with 
$$\int \coth^2 t \,dt=t-\coth t +C.$$
(It helps to know the derivatives of the hyperbolic functions, and some identities.) 
A: In the linked graphic you posted, the author demonstrates that the first integral is greater than a second integral (namely $2\pi\int ydx$), and the second integral diverges, hence the surface area is infinite (no finite quantity may be greater than an infinite one). This is possible because $$y\sqrt{1+(y')^2}>y,$$ and inequalities are preserved under integration (loosely speaking). This is the comparison test.

I'm doubtful you actually have to find the exact value of $$\int_1^a \frac{1}{x}\sqrt{1+\frac{1}{x^4}}dx,$$ for unknown $a$, as I can't do it off the top of my head and WolframAlpha stutters on it. If that is indeed what you are being asked to do then feel free to update us.
