# Area vs Volume Paradox [duplicate]

Hi,

I have this question that I quite cant explain why.

So the area under the curve $$y=\frac{1}{x}$$ from 1 to infinity equals $$\lim_{r->\infty } \ln(r) |_1^r = \infty$$

And the volume of the solid by rotation the equation about the x-axis from 1 to infinity is

$$\int_{1}^{\infty} \pi\left (\frac{1}{x} \right )^2 dx$$ which evaluates to the area being $$\pi$$

So, whats the explanation for the area being infinite but the volume being finite?

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## marked as duplicate by Qiaochu YuanJun 1 '11 at 22:04

You can accomplish a similar situation if you imagine starting with a cylinder of, say, radius $r$ and height $h$. The volume is $\pi r^2h$, the surface area is $2\pi rh$. If try to make the cylinder taller and thinner so as to keep the volume constant, you will necessarily increase the height a lot faster than you are decreasing the radius, because the decrease in the radius will be quadratic while the increase in the height will be linear (if you decrease the radius by a factor of $\frac{1}{4}$th, the volume decreases by a factor of $\frac{1}{16}$, so you would need to increase the height by a factor of $16$ to keep up); however, that will result in a net increase in the surface area (if you decrease the radius by $\frac{1}{4}$ and increase the height by a factor of $16$, then the surface area increases by a factor of $4$). If you keep making the cylinder taller and thinner and keeping the volume constant, then the surface area will necessarily grow without bound. So you have a "continuous deformation" of the cylinder that keeps the volume constant, but the surface area grows without bound.
It's just like the fact that you can add an infinite number of positive numbers, $$\frac{1}{1^2} + \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots+\frac{1}{n^2} + \cdots$$ and yet get a finite total, but if you change it just a bit you can get something infinite: $$\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} +\cdots$$ which diverges.