# Thread: Conceptualising integration/vol. of revolution

1. ## Conceptualising integration/vol. of revolution

The area under the curve y= x^(-2/3) from x=1 to infinity is undefined. Yet if the region is rotated about the x-axis the volume of revolution IS defined as 3pie.
I've done the math but can't explain how it is so conceptually. The best I can come up with is that the area is not necessarily infinite just indeterminable. But I think that this may be b.s.
Any help would be greatly appreciated.

Woollybull.

2. Why not take a shot at the surface area, while you are at it?

"Conceptually"? What does that mean? The "concept" you should be contemplating is "convergence". It converges or it doesn't.

Perhaps the answer lies in your use of the word "pie". Pie is for eating. Pi is for mathematics and Greeks. Are you distracted by lunch and not really thinking about mathematics?

Also, your use of the word "infinite" is not useful. There may be a day, in Geometry or Complex Analysis, where "infinity" will be a place to stop and visit, but if you abandon this for now and consider simply that the value of the integral increases without bound as x increases, where's the question?

The physical concept of NONconvergence is very simple:

1) Pick a value.
2) You will exceed it eventually.

You see, it is not a matter of being "indeterminable". It is a matter of being able to demonstrate that there is no such value to which the integral settles.

3. I am unable to visualise how the area doesnt settle but the volume does settle. Can you help me with that??

Thanks.

4. Surface area is infinite, but volume is finite.

$2{\pi}\int_{1}^{\infty}(\frac{1}{x^{\frac{2}{3}}}) \sqrt{1+\frac{4}{9x^{\frac{10}{3}}}}dx$

But, $\sqrt{1+\frac{4}{9x^{\frac{10}{3}}}}\geq{1}, \;\ if \;\ x\geq{1}$

So, $\frac{2{\pi}}{x^{\frac{2}{3}}}\sqrt{1+\frac{4}{9x^ {\frac{10}{3}}}}\geq{\frac{2{\pi}}{x}}$

Therefore, $\lim_{L\rightarrow{\infty}}\frac{2{\pi}}{x}dx={\in fty}$

Think of it as filling the solid with paint and letting it seep through the surface.
Then you could paint an infinite surface area with a finite amount of paint.