Sphere

Agent Smith

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Oct 18, 2023
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So just learnt, for

1. \(\displaystyle V = \frac{4}{3} \pi r^3\), \(\displaystyle \frac{dV}{dr} = 4\pi r^2\)

2. \(\displaystyle A = \pi r^2, \frac{dA}{dr} = 2\pi r\)

What is the significance of these relationships?
 
So just learnt, for

1. \(\displaystyle V = \frac{4}{3} \pi r^3\), \(\displaystyle \frac{dV}{dr} = 4\pi r^2\)

2. \(\displaystyle A = \pi r^2, \frac{dA}{dr} = 2\pi r\)

What is the significance of these relationships?
1. Imagine increasing a sphere by a small amount, [imath]dr[/imath] (e.g. by painting it). The change in volume is the thickness of the paint times the surface area: [imath]dV=Sdr=4\pi r^2dr[/imath].

2. Imagine increasing a circle by a small amount, [imath]dr[/imath] (e.g. by drawing a thin layer around it). The change in area is the thickness of the layer times the circumference: [imath]dA=Cdr=\frac{4}{3}\pi r^3dr[/imath].

Not all shapes work this way, but these and a couple more do.
 
Not all shapes work this way, but these and a couple more do.
Somewhat tangential, but this can be generalized to all smooth surfaces (and even piece-wise smooth) if one changes from [imath]dr[/imath] to [imath]dn[/imath], i.e. shifting the surface by an infinitesimal amount along its normals, i.e. [imath]dV = Sdn[/imath]. This is easy to verify, for example, in the case of rectangular cuboids.
 
@Agent Smith, be a gentleman and tell us frankly, do you agree with this:

[imath]\displaystyle dA = C dr = V dr[/imath]
I don't see a choice in the matter. It makes sense to me - these explanations above. To note is Chevalieri's principle; however I don't know how applicable it is to the issue. Do you have other explanations?
 
I don't see a choice in the matter. It makes sense to me - these explanations above. To note is Chevalieri's principle; however I don't know how applicable it is to the issue. Do you have other explanations?
If you see something wrong, you have to say something about it even if it came from Einstein. Do you think that it is logical to say the circumference of a circle is equal to the volume of a sphere?
 
If you see something wrong, you have to say something about it even if it came from Einstein. Do you think that it is logical to say the circumference of a circle is equal to the volume of a sphere?
Ok, but aren't we, it looks like, we're talking about concentric circles/spheres, oui? Confetior, non liquet.
 
1. Imagine increasing a sphere by a small amount, [imath]dr[/imath] (e.g. by painting it). The change in volume is the thickness of the paint times the surface area: [imath]dV=Sdr=4\pi r^2dr[/imath].

2. Imagine increasing a circle by a small amount, [imath]dr[/imath] (e.g. by drawing a thin layer around it). The change in area is the thickness of the layer times the circumference: [imath]dA=Cdr=\frac{4}{3}\pi r^3dr[/imath].

Not all shapes work this way, but these and a couple more do.
Concentric circles/spheres, mon ami?

Is it possible to improve upon your intuition? Much obliged.
 
it seems that he believes of anything the seniors say
💡 Be careful when judging replies in which an entire post has been quoted (as seen in post #10). It's a possibility that Agent Smith's statements in post #10 refer only to Dr. Peterson's second line. (It could very well be that Agent Smith has ignored the first line.)

The issue here is that most forum members do not properly quote posts, when they reply only to a specific part. In such cases, the proper way is to quote only what you intend to post about. The lazy way is to simply quote the entire post, often leaving the audience to determine what specifically in the quote the reply is about. (By default, the forum software does not display long quotes in their entirety. Such quotes frequently need to be unnecessarily opened, whereas most proper quotes would be displayed without having to open them and hunt.)

For all readers: When using a computer, it's quite easy to quote properly. Simply highlight the statement(s), and click the pop-up reply link. (I don't recommend doing this while viewing the forum's mobile site; the pop-up link is often dead.) 💡
[imath]\;[/imath]
 
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