We are asked to confirm that the volume of the figure known as Gabriel's Horn is finite.

We will use the fact that ` int_1^( oo) (dx)/x^p={[[1/(p-1),"if" p>1],["diverges", p<=1]] `

The solid is generated by revolving the unbounded region between the graph of `f(x)=1/x ` and the x-axis, about the x-axis for `x>=1 ` .

We use the disk method: each disk is a circle of radius f(x).

`V=pi int_1^(oo) (1/x)^2dx `

`V=pi int_1^(oo) (dx)/(x^2) `

Using the Lemma above we get:

`V=pi(1/(2-1))=pi ` which of course is finite.

**Further Reading**

Gabriel's Horn is made by revolving the function `f(x)=1/x` about the x-axis, with the domain `1lt= x` . The volume can be found by slicing the horn up into infinitesimal circles of radius `f(x)` . Then summing their areas from `1` to `oo` .

`V=int_1^oo pi r^2 dx=pi int_1^oo f(x)^2 dx `

`V=pi int_1^oo 1/x^2 dx=pi(-1/x)|_1^oo=pi [(lim_(x-gtoo)-1/x)+1/1]`

`V=pi(0+1)`

`V=pi`

As you can see the volume of the horn is exactly `pi` .

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