Vol. & Area of Revolved Graph: Prove Finite Volume & Infinite Area

[cooltee13]

Homework Statement

Graph 1/(x^2), and revolve it around the x-axis to form a 'horn' type shape. Prove that the volume is finite, while the area is infinite

Homework Equations

no specific equations.. I know that to find the volume you need to use the shell method and take the sum of all the little 'shells'. Also, the area of a cylinder is pi*r*h, but I am not sure how to prove that its finite, while the area is infinite.

The Attempt at a Solution

Well like i said, basically I tried to calculate the volume: V = the sum pi(Xi^2-Xi-1^2)(Si-Si^2) where Xi -Xi-1 is the width of the cylinder, and Si-Si^2 is the height. But like i said, that's where i get lost.. I am pretty sure that the answer is going to be something like 1/infinity for the volume and 1/0 for the area..but I am not positive.

Any help would be appreciated, Thanks !

 

[dynamicsolo]

Homework Statement

Graph 1/(x^2), and revolve it around the x-axis to form a 'horn' type shape. Prove that the volume is finite, while the area is infinite

I believe the specification of this problem is incomplete and/or incorrect. The "horn" which would have finite volume but infinite area is the figure of revolution for y = 1/x.

But there is another difficulty here. Are you supposed to use an improper integral to find the area and volume? You will not be able to get the volume integral to converge if the curve is to run from x = 0 to x = infinity. I suspect it is intended that you start the curve from some non-zero value of x (typically x = 1) and extend the curve "to infinity". But y = 1/(x^2) will give you a "horn" with finite volume and area in that case.

 

[cooltee13]

Ahh, I just talked with a classmate, and its 1/sqrt(x) not 1/(x^2) my apologies. You do have to use a volume integral, but I am not very familiar with the steps.

 

[dynamicsolo]

Ahh, I just talked with a classmate, and its 1/sqrt(x) not 1/(x^2) my apologies. You do have to use a volume integral, but I am not very familiar with the steps.

There is still the issue of the integration limits. If the person who posed this problem didn't state any, I suppose we could integrate from x = 1 to x = infinity. Have you covered improper integrals yet?

P.S. You're sure it's "1/sqrt(x)"? I'm not sure the volume integral will converge either for that function. (The classic problem of this sort is to use the function 1/x .)

 

[cooltee13]

Well, Like i said, A classmate thought it was 1/sqrt(x) and i thought it was 1/x^2..but we both could be wrong i suppose. I graphed 1/x and it does seem to make more sense, so Ill just go with that and If its not correct so be it.

He didnt state any integral either, just the function and the question stated. I've covered improper integrals before yes, but It was last semester so I may be a little rusty.

 

[cooltee13]

-This is what I've figured out so far:

The area of the curve under the graph can be approximated using calculus, hence its finite. And if I revolve it, the volume is finite, because the area used was finite. However To get the surface area I would need to rotate the length of the curve, which is inifinite, so the SA is infinite as well.

I know this, but I just can't figure out a way to prove it mathematically

 

[dynamicsolo]

-This is what I've figured out so far:

The area of the curve under the graph can be approximated using calculus, hence its finite. And if I revolve it, the volume is finite, because the area used was finite. However To get the surface area I would need to rotate the length of the curve, which is inifinite, so the SA is infinite as well.

I know this, but I just can't figure out a way to prove it mathematically

The area under 1/x is not finite if you are going out to infinity: you would integrate 1/x from, say, 1 to infinity, which leads to evaluating

(lim x->inf. ln x) - (ln 1) .

The volume is obtained in this way. Imagine an infinitesimal "slice" of area with height 1/x and width dx. When you revolve it around the x-axis, you now have a disk of area
(pi)·[(1/x)^2] and thickness dx. Upon integrating these infinitesimal volumes from 1 to infinity, you would evaluate

(pi)·{ (lim x->inf. [-1/x] ) - ( -1/1 ) } ,

which is finite.

The surface area of this horn is found by integrating the revolved infinitesimal bits of arclength along the curve, which are infinitesimally-wide bands having width dx and circumference

(2·pi) · [ 1 + (dy/dx)^2 ]^(1/2) .

So you would integrate (2·pi) · [ 1 + {1/(x^2)} ]^(1/2) ] dx from 1 to infinity,
which turns out not to be finite.

EDIT: Your statement about the arclength of the curve is essentially correct. The surface area of the curve is found by integrating a term larger than 1 from 1 to infinity. Since the integral of (1 dx) from x = 1 to x = infinity won't converge, integrating a term larger than 1 certainly won't converge either. So the surface area of such "power-law" horns is infinite, but the volume for a horn using a power-law 1/(x^p) with p > (1/2) will have finite volume.

When I was first learning about integrating "solids of revolution", we were shown a short cartoon involving someone who'd been set the task of painting a house which was built as half of the "horn" produced by 1/x . They learn that they can't do it because, while the volume of the house is finite, the surface of the house isn't. That still amazes me (as a physicist) in that you could fill the house with a finite amount of paint, but you can't coat it. (Such are the nature of mathematical infinities.)

 

[cooltee13]

ahh ok, that makes sense, well, ill just go with what we discussed..hopefully i chose the correct function :/

thanks for your help though