B "Onion proof" of the area of a circle

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K41

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I understand the basic concept, although it is a little difficult to visualize the thin discs close to the centre of the circle. Regarding the area of each disc, it is given in the link above as 2πrdr. Then, by means of integration, each disc is "added" and the total area of the circle obtained.

Is it valid to think of each thin disc as a rectangle of length "2πr" and height "dr"?
 

mathman

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Thinking it as a rectangle is unnecessary, but if it works for you - OK.
 
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Of course it is, were you to "unroll" it, it'd be just that.
 
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I understand the basic concept, although it is a little difficult to visualize the thin discs close to the centre of the circle. Regarding the area of each disc, it is given in the link above as 2πrdr. Then, by means of integration, each disc is "added" and the total area of the circle obtained.

Is it valid to think of each thin disc as a rectangle of length "2πr" and height "dr"?
I suspect that was a abuse of language; the way I learned, infinitesimals of higher order (like ##(\Delta x)^2## ) vanish when compared to infinitesimals of first order (like ##\Delta x##). In the figure below:
241051


The area ## \Delta A ## is a segment of a ring with thickness ##\Delta r## and inner radius ##r \Delta\phi## and outer radius ##(r + \Delta r) \Delta\phi##; now, for ##\Delta\phi## very small,

##sin \Delta\phi = \Delta\phi + \epsilon (\Delta\phi)^3 + ...##

when the limit is taken, the term ##(\Delta \phi)^3## becomes an infinitesimal of higher order, so it will vanish when compared to ##\Delta\phi##; therefore ##sin \Delta\phi = \Delta\phi## when the limit is taken. That is, when the limit is taken the inner radius is the same as ##r sin \Delta\phi##, that is, the same as the base of a trapezoid.

Now, the outer radius of the ring is ##(r+\Delta r)*\Delta\phi##, which is the same (on the limit) as ##(r + \Delta r) * sin \Delta\phi##, which is the same as the second base of the trapezoid.

It so happens that ##(r + \Delta r) * sin \Delta\phi = r.sin \Delta\phi + (\Delta r)*(sin \Delta \phi) = (r.sin \Delta\phi )+ (\Delta r)*( \Delta\phi )## and ##(\Delta r)*(\Delta\phi)## is also an infinitesimal of higher order, so it will vanish too on the limit; therefore, the length of the second base is the same as the first base, making that a rectangle on the limit, not a trapezoid. The rectangle has base ##(r.sin \Delta\phi)## and height ##\Delta r##. I think that's why the guy wrote that the infinitesimal ring segment is equivalent to an infinitesimal rectangle, or abusing the language, the ring is equivalent to a rectangle.

So ##\Delta A = (\Delta r)*(r.sin \Delta\phi)## + (higher order infinitesimals) = ##r \Delta r \Delta\phi## + (higher order infinitesimals), and

##lim_{r->0,phi->0} \Delta A = d^2A = lim_{r->0,phi->0} (r \Delta r \Delta\phi) + lim_{r->0,phi->0}## (higher order infinitesimals)## = r.dr.d\phi + 0##

integrating over ##\phi##, the area of the ring is then ##dA = 2 \pi r dr##

Keep in mind, I'm just a low level Padawan here, so someone will probably admonish me for abusing notation too.. :)
 
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Err... typo... those limits are on ##lim_{\Delta r ->0,\Delta \phi ->0}##, not ##lim_{r ->0,\phi ->0}## ... haha... I'm so bad on writing properly :D
 

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