# Surface area of a solid of revolution

1. Mar 9, 2009

1. The problem statement, all variables and given/known data

Consider now a shape that is obtained by revolving
the “inﬁnitely long” function
f (x) = 1/x , 1 ≤ x < ∞
around the x-axis. Find both the surface area of the resulting object, and the
enclosed volume of it, i.e. the volume of the solid obtained from revolving the
area under the graph of f (x) around the x-axis.

If you do this correctly, you will ﬁnd an inﬁnite surface area but a ﬁnite volume. This would mean that a certain (ﬁnite) amount of paint is suﬃcient to ﬁll up our object, but that this
amount would not be enough to actually paint it. Does this surprising result
mean that there is something fundamentally wrong with our approach? Why
or why not? Comment critically!

2. Relevant equations
I understand how to calculate the surface area and the volume of a solid of revolution, however it's the second part of the question I don't really undertand, the part about there being something fundamentally wrong with our approach. I was wondering if anyone could give me some hints to point me in the right direction.

2. Mar 9, 2009

### Staff: Mentor

A good start would be to actually calculate the surface area and volume for this solid. Keep in mind that you are going to have improper integrals for both, since one of the limits of integration is $\infty$.

3. Mar 10, 2009