Question

Suppose f(x) is continuous on [a; b] and is revolved about the x-axis. The surface area of the resulting figure is given by

integral from a to be of 2(pi)y * sqrt(1+(dy/dx)^2)

-Consider the function f(x) = 1/x on the interval [1; infinity). Show that the surface area of f revolved about the x-axis on the same interval is infinite.

Answer 1

have you set up the integral yet?

Answer 2

\[\int\limits_{1}^{\infty}\frac{ 1 }{ x }dx\]
this one?

Answer 3

what happened to the sqrt term?

Answer 4

you mean this one?
\[\int\limits_{a}^{b}2 \pi y \sqrt{1+(\frac{ dy }{ dx }})^2\]

Answer 5

yeah, but use y=1/x..

Answer 6

so does it become
\[\int\limits_{1}^{\infty}2 \pi (\frac{ 1 }{ x })\sqrt{1+(\frac{ -1 }{ x^2 }})^2\]

Answer 7

yes, now my advice is to simplify it

Answer 8

so:
\[2 \pi \int\limits_{1}^{\infty}\frac{ 1 }{ x }\sqrt{1+(\frac{ 1 }{ x^4 }})\]

Answer 9

right, now I'm not so sure of the best route from here, but it seems to me a good idea to pull 1/x^4 out of the radical

Answer 10

you should really be writing dx, it is important at this point in calc

Answer 11

oh yes ill remember to add dx! also should i try finding the limit as say "t" approaches infinity and make the integral go from 1 to "t"?

Answer 12

yeah, but you needn't write that until you are about to evaluate.
for now it seems to me that you want to do this with a somewhat tricky trig substitution. If there is a better way I don't see it yet.

Answer 13

but first I would factor 1/x^4 out of the radical as I said

Answer 14

so would it then become \[1/x^4\sqrt{x^4+1}\]

Answer 15

then all together \[2 \pi \int\limits_{1}^{\infty}1/x^5\sqrt{x^4+1}\]

Answer 16

you factored that wrong...

Answer 17

\[\sqrt{1+\frac1{x^4}}=\sqrt{\frac {x^4+1}{x^4}}=\frac1{x^2}\sqrt{x^4+1}\]

Answer 18

ah i see..

Answer 19

Do you know about hyperbolic functions? Because I doubt that these kind of substitutions can be avoided here.

Answer 20

I am thinking that what might be better is to do this along y instead

Answer 21

If it can be simplified by changing a different method it's even better, the above integral is pretty complex (-:

Answer 22

let \(x=1/y\), then as \(x\to\infty,~~y\to0\)\[S=2\pi\int_1^0 y\sqrt{1+\frac1{y^4}}dy\]

Answer 23

not sure though, just a thought

Answer 24

still a nasty integral...

Answer 25

I thought I should write down some of the imperative steps to solve this integral, for completeness reasons of this post, however I would like the opener @peggiepenguin to make sure that we're talking about the same problem here. However:

\[\Large \int \frac{1}{x^3}\sqrt{x^4+1}dx \]

Setting \(x^2=\sinh \alpha\) such that \(dx/d\alpha=\cosh\alpha/(2\sqrt{\sinh \alpha}) \)
Substituting this back yields to:
\[\Large \int\frac{1}{\sqrt{\sinh^3\alpha}} \cdot\frac{\sqrt{\sinh^2\alpha+1}}{2\sqrt{\sinh\alpha}} \cdot\cosh\alpha d\alpha\]

Hyperbolic trigonometry:
\[\Large \cosh^2\alpha-\sinh^2\alpha=1 \longrightarrow \cosh^2\alpha=\sinh^2\alpha+1 \]
Cleaning this all up:
\[\Large \frac{1}{2}\int \frac{\cosh^2\alpha}{\sqrt{\sinh^4}\alpha}d\alpha=\frac{1}{2} \int\frac{\cosh^2\alpha}{|\sinh^2\alpha|}d\alpha\]

Cheating a little, considering the numerator as positive within the domain of the integral:
\[\Large \frac{1}{2}\int\frac{\cosh^2\alpha}{\sinh^2\alpha}d\alpha=\frac{1}{2}\int\coth^2\alpha d\alpha =\frac{1}{2}(\alpha-\coth\alpha)\]

Answer 26

Trigonometric back-substitution:
\[\Large \sinh\alpha=x^2=\frac{x^2}{1} \]

and remember the hyperbolic Pythagoras
\[\Large \cosh^2\alpha-\sinh^2\alpha=1 \\ \Large x^2-y^2=r^2 \]

So: