# 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 $$\int 2\pi y dx$$

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In the linked graphic you posted, the author demonstrates that the first integral is greater than a second integral (namely $2\pi \int y dx$), 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 – anon Aug 18 '11 at 3:27
Thank you for your concise and clear response thank you very much anon. – FlightOfGrey Aug 18 '11 at 3:55
The antiderivative could be calculated explicitly, but it is kind of hard. If you did it, you would find that the definite integral diverges (is "infinite"). But a simple inequality enables us to bypass the ugly integration. – André Nicolas Aug 18 '11 at 4:23
@anon, perhaps you'd consider posting your comment as an answer? – Gerry Myerson Aug 18 '11 at 10:19

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.)

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As a tiny aside: the mess of logarithms before $\frac1{t}$ in the final result can be replaced with $\mathrm{artanh}(t)$ if wanted. Alternatively, since $1+(it)^2=1-t^2$ and the integral of $\frac1{1+t^2}$ is slightly more well known... :) – J. M. Aug 19 '11 at 0:54
@J. M.: I typed in an addendum on how to use hyperbolic function substitution in the integration. – André Nicolas Aug 19 '11 at 1:11
This is what I was looking for, I am actually learning Hyperbolic functions at the moment in my first year calculus course, however I don't completely understand them yet myself and that is probably the answer that the markers were looking for. Why did you choose to let $x^2 = tan\theta$ in your first answer and $x^2 = sinh t$ in your second? – FlightOfGrey Aug 20 '11 at 0:35
@FlightofGrey: Thought if I could take the square root of $1+x^4$, everything might be OK. We have $1+\tan^2\theta=\sec^2\theta$, so if we let $x^2=\tan\theta$, we'll be able to take square root of $1+x^4$, it will be $\sec\theta$. Similarly, $1+\sinh^2 t=\cosh^2 t$, which means that if $x^2=\sinh t$, we can take square root of $1+x^4$. And sine and cos, hyperbolic or not, are often nicer. Anything that makes square root possible is a good idea. Not all plausible things to try work out, unfortunately. Had a good feeling about $2/x^3$, odd powers are nicer than even. – André Nicolas Aug 20 '11 at 1:10

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.

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We have to compute the surface area from $x=1$ to $x=a$, where $a$ is any real number greater than 1 and then compute $\lim_{a\to\infty}Surface Area$ – FlightOfGrey Aug 20 '11 at 0:27