Excellent question: A=infinity, V=pi?

[silverdiesel]

My professor posed a brain teaser question today, and I can't get it out of my mind. I was hoping the forum can help me make sense of it.

Area under 1/x = infinity:
$$A = \int_{1}^{\infty} (1/x)dx$$
$$A = \lim_{t\rightarrow \infty} \int_{1}^{t} (1/x)dx$$
$$A = \lim_{t\rightarrow \infty} \ln t - \ln 1$$
$$A = \infty$$

but the volume of 1/x rotated around the x-axis is equal to pi
$$V = \pi \int_{1}^{\infty} (1/x)^2dx$$
$$V = \pi \lim_{t\rightarrow \infty} \int_{1}^{t} (1/x)^2dx$$
$$= \pi \lim_{t\rightarrow \infty} (-1/t + 1/1)$$
$$= \pi (1)$$

How can this be true?

 

[0rthodontist]

Well, one way to break down your intuitive opposition to it is that for circles with small radii, pi times the square of the radius is smaller than the radius itself.

 

[TD]

How can this be true?

Why can't it be true ?

It seems that popularly said, it is perfectly possible to create a mathematical 'object' with a finite volume but an infinite surface area (which is 2*pi times the integral you calculated first).

 

[AKG]

See .[/URL] Especially the paragraph about the "paradox" that you could fill this shape with a finite amount of paint, but cover an infinite surface area.

 

[benorin]

The solid of revolution formed by rotating y=1/x about the x-axis from x=1 to infinity is called Gabriel's Horn.