1
$\begingroup$

The integral of the area of a circle ($\pi{r^2}$) is $\frac{1}{3}\pi{r^3}$, which is the volume of a cone when its height is equal to the radius of its base. It is my understanding that, in-keeping with the whole idea of calculus, you can view this relationship as infinitesimal circle areas, each with an infinitesimally smaller radius than the last, being stacked up on top of each other, forming a conic shape.

The original area was two dimensional, and through integration, it has been projected into three dimensions.

So, my question is, does the integral of the volume of a cone, which is $\frac{1}{12}\pi{r^4}$, describe the volume (or equivalent) of a shape composed of infinitesimally shrinking/enlarging cones being stacked on top of one another, projected in the fourth dimension?

$\endgroup$
  • $\begingroup$ Dear Kris Walker--why did you delete your good question on cylindrical helices? Bummer! I was having such a good time answering it . . . $\endgroup$ – Robert Lewis Aug 19 '18 at 1:08
  • $\begingroup$ I am so sorry Robert. I copied the question from my assignment and felt like I shouldn't keep it up. Is your work gone? $\endgroup$ – Kris Walker Aug 19 '18 at 1:13
  • $\begingroup$ No, I saved it in a local file. I'm almost tempted to ask a similar question now myself; did you get an answer? $\endgroup$ – Robert Lewis Aug 19 '18 at 1:15
  • $\begingroup$ Oh good. I didn't, no, but I think I'd like to figure it out on my own anyway. I'd love to see your answer though so perhaps you should :) $\endgroup$ – Kris Walker Aug 19 '18 at 1:18
  • $\begingroup$ OK, I'll probably re-post it on my own and let you know when I have something. Probably it will take long enough that you'll have figured it out by then! $\endgroup$ – Robert Lewis Aug 19 '18 at 1:21
0
$\begingroup$

Well, according to this Wiki page:

$$\text{V}_\text{n}\left(\text{R}\right)=\frac{\pi^\frac{\text{n}}{2}}{\Gamma\left(1+\frac{\text{n}}{2}\right)}\cdot\text{R}^\text{n}\tag1$$

So, when $\text{n}=4$:

$$\text{V}_4\left(\text{R}\right)=\frac{\pi^\frac{4}{2}}{\Gamma\left(1+\frac{4}{2}\right)}\cdot\text{R}^4=\frac{\pi^2}{2}\cdot\text{R}^4\tag2$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.