Question

Find volume or state it does not exist

Region bounded by f(x) = sqrt((x+1)/(x^3)) and the x axis on the interval [1,infinite) is revolved about the x axis

I though of using the shell method

so I got this far 2pi integral from 1 to infinite x(sqrt((x+1)/(x^3)))

Maybe the disk method would be easier...any thoughts?

Either way I am stuck here.

Answer 1

@max0236 how did you writ this >>> \(\Huge{}\)

Answer 2

\[2\pi \int\limits_{1}^{\infty} x \sqrt{(x+1)/x^3}dx\]

Answer 3

try discs:

Answer 4

you get a finite volume.

Answer 5

questions?

Answer 6

I finally got \[\frac{ 2b+1 }{ 2b^2 }- \frac{ 2(1)+1 }{ 2(1)^2 } \]

now I substituted in b for infinite. So i plugged in b and 1 to the function. So how do I know its finite

Answer 7

the limit is n-> infinity (-2 +1/n) / n
aka zero :)

Answer 8

whoops left out a '2' :

bit off... the integral is - (2x+1)/2*x^2

so : { lim n->inf of -(2n+1)/n^2 } - - (2(1) +1)/(2*1^2) (all multiplied by pi)

Answer 9

right a number over infinite goes to zero I recall now. Thank you very much!

Answer 10

sure:)