Paradox? An infinite set having finite volume?

[mrdoe]

Paradox? An infinite set having finite volume??

Homework Statement

Find the volume, in terms of k, of the solid made from R rotating about the y-axis, if R is defined as the region bounded by e^{-x} and the coordinate axes.
2. The attempt at a solution
Obviously, \displaystyle\int^{1}_{0} (\ln (y))^2\pi dx = undefined... but the region R itself has a finite volume. So if you rotate it shouldn't it have a finite volume?

 

[CompuChip]

I assume that you mean
<br /> \displaystyle\int^{1}_{0} (\ln (y))^2\pi dy =
because you are integrating over y.

The integral can be done, but it is not trivial to find the primitive.

My suggestion would be to "try"
y \ln^2(y)
which upon differentiation gives you in any case the integrand back... then try to cancel out the unwanted term by adding something.

As for the evaluation of the integral, it looks like it is diverging. However, if you take an appropriate limit, you will still get the correct (finite) answer. The trick here is that your logs will only occur in terms like
y \ln y, y \ln^2 y, \cdots
and that the linear term goes to zero faster than the logarithm goes to infinity, as y -> 0.

 

[Dick]

You don't really have to guess the primitive. The systematic way to do the integral is to do integration by parts twice. Start with u=ln(y)^2 dv=dy.

 

[mrdoe]

Yeah I'm getting the following for indefinite integral...

\pi x\ln^2 x - 2\pi x\ln x + 2\pi x

But then if you try that with 0 it's undefined so... would the answer be 2\pi?

 

[Dick]

CompuChip already said this, but I'll repeat. Things like x*ln(x) may be undefined at x=0, but they do have a limit as x->0. You do have to check that limit is 0 before you can ignore those terms. You can't ignore something just because it's 'undefined'.