Gabriel's horn is formed by revolving the curve $y=1/x$ for $x\in [1,\infty)$ about the $x$-axis.
Find the volume inside Gabriel's horn. I have the answer but I can't seem to get it right. Can someone explain please?
[Math] Gabriel’s horn, calculus Integration
calculusintegration
Related Solutions
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.)
Well, first one has the following fact:
If one makes the corresponding "Gabriel's Horn" by rotating the region under $\frac{1}{x^p}$ from $x=1$ to $x=\infty$ around the $x$-axis, then:
The volume is finite iff $p>\frac{1}{2}$
The surface area is finite iff $p > 1$ so in particular
The Horn has finite volume and infinite surface area for any $p$ between $\frac{1}{2}$ and $1$, where $p$ cannot be $\frac{1}{2}$ but it can be $1$.
The proof is simply several uses of the comparison theorem.
One cool thing that pops out is that the standard Gabriel's horn $(p=1)$ actually gives the minimal volume Gabriel's horn which still has infinite surface area. If you shrink the volume at all (i.e., $p$ goes up), then the surface area becomes finite.
There are, of course, variations on this idea. For example, one can add some small bumps to the graph of $\frac{1}{x}$. Note that bumps will tend to increase the surface area (so it will stay infinite), but if the bumps are not too frequent or too high, then the volume will still converge. (Concretely, imagine rotation $\frac{\sin x}{x}$ with $1\leq x < \infty$ around the $x$ axis).
Finally, there are some other shapes which are in no way related to Gabriel's horn. For example, the Menger Sponge has finite volume but infinite surface area. You start with a cube of volume $1$ (so, very finite), and start cutting pieces out. Of course, this only lowers the volume, so the volume is finite at the end of the day. On the other hand, cutting pieces out increases the surface, and at the end up the day it's infinite.
Best Answer
What you should do for finding the result is to evaluate the following integral $$\int_1^\infty \pi y^2 dx$$ in which $y=\frac{1}{x}$. You can draw a disc as you see in fig below. We build this disk on $x$- axes, so the volume of it is the volume of colored cylinder. What is that volume? It is $\pi r^2 h$. What is $r$ and what is $h$? indeed, $r$ is $y$ and $h$ is $dx$.