Is my reasoning for calculating the area of a ring using calculus correct?

[nobahar]

Hello!
This is probably a really asinine question.
I was trying to identify an area of a ring, namely a really small ring such that its near enough the circumference of a circle. I thought I could approach it in two ways.
The first was to subtract a smaller circle of radius r₁ form a circle of larger radius r₂:
\pi r_{2}^{2} - \pi r_{1}^{2} = \pi (r_{2}^{2} - r_{1}^{2})

Using the limit r₁ tends to r₂, this should be a ring with an area equal to that which I could obtain from using the circumference of a circle of radius r₂ multiplied by a tiny amount (essentially giving it some thickness:
This is only approximately a ring:
2\pi r_{2} (r_{2}-r_{1})

But in the limit it would be a ring:
2\pi r_{2} \left dr

Are the two equal? Conceptually, it seems they should be.
This was my reasoning:
\pi r_{2}^{2} - \pi r_{1}^{2} = \pi (r_{2}^{2} - r_{1}^{2}) = \pi (r_{2} + r_{1})(r_{2} - r_{1})

In the limit:
r_{2} - r_{1} Approaches 0, but is some infinitesimally small difference dr, which is explicitly non-zero.
The summation component:
r_{2} + r_{1} Seems to approach 2.

So we have:
\lim \pi (r_{2} + r_{1})(r_{2} - r_{1}) = \pi(2r_{2}}{dr}

Is it considered to equal 2 or is it explicitly not 2 as the difference component is explicitly not 0?
I am confused, is there an error in my reasoning?
Any help appreciated as this is very frustrating!
Thanks in advance,
Nobahar

 

[rock.freak667]

Well your area would be A=2πrt where t is the thickness.

If you consider the ring with thickness 'dr', the area is 2πr*dr, so the area is A=∫2πr*dr from r₁ to r₂ which gives the same thing.

 

[tiny-tim]

hello nobahar!

isn't this begging the question?

you want to prove that the area of a circle is πr²,

and you're trying to prove it by using the area of a ring which assumes the formula you're trying to prove

instead, just say that the circumference of the tiny ring is 2πr (that's the definition of π ), and its width is dr, so its area is 2πrdr … and then integrate that!

 

[nobahar]

Hello!
Thanks for the responses! I apologise for not replying sooner as I have had exams recently (damn exams, always interfering with my studies!).
Anyway, I was trying to obtain a ring, a ring of the most minutest size. Essentially the circumference of a circle.
It related to adding up the force all exerted on an object. The forces all emanate from a ring, and the object which expreiences the forces is susended above the centre of the ring. The horizontal components attributed to each point on the ring cancel, and the vertical component is all that remains. Lots of these rings are then added up to give a total flat surface area, its just easoer to do the calculation using rings.
With this in mind. I needed the force exerted from a ring of very small thickness, dr. I was confused, as I thought I could approach it from either direction. First, using the equation for the circumference of a circle and extending it to a thickness of dr. Alternatively, by using two circles, and subtracting one from the other; making the second, subtracted circle almost exactly the same size as the first. Differing in radius by an amount dr.
Conceptually, I don't see why the shoudl not arrive at the same conclusion.
Thanks for your input tiny and rock.

 

[HallsofIvy]

Hello!
This is probably a really asinine question.
I was trying to identify an area of a ring, namely a really small ring such that its near enough the circumference of a circle. I thought I could approach it in two ways.
The first was to subtract a smaller circle of radius r₁ form a circle of larger radius r₂:
\pi r_{2}^{2} - \pi r_{1}^{2} = \pi (r_{2}^{2} - r_{1}^{2})

Yes,this is correct.

And, now, you want to assume that r_2 and r_1 are very close: that r_2= r_1+ dr for some very small "dr". Then r_2^2= (r_1+ dr)^2= r_1^2+ 2rdr+ (dr)^2. Since dr is very small, its square is much smaller and, comparatively, we can ignore it: approximately, at least, r_2^2= r_1^2+ 2rdr and so r_2^2- r-1^2= 2rdr. Using that, your area is \pi(r_2^2- r_1^2)~ \pi(2r dr)= (2\pi r)dr.

2\pi r is the circumference of the ring so that is the formula you want: circumference times thickness. That also says, by the way, that you could "cut" the ring at one point, the "straighten it out" to get the area as a rectangle- the length (circumference) time the width (thickness).

Using the limit r₁ tends to r₂, this should be a ring with an area equal to that which I could obtain from using the circumference of a circle of radius r₂ multiplied by a tiny amount (essentially giving it some thickness:
This is only approximately a ring:
2\pi r_{2} (r_{2}-r_{1})

But in the limit it would be a ring:
2\pi r_{2} \left dr

Are the two equal? Conceptually, it seems they should be.
This was my reasoning:
\pi r_{2}^{2} - \pi r_{1}^{2} = \pi (r_{2}^{2} - r_{1}^{2}) = \pi (r_{2} + r_{1})(r_{2} - r_{1})

In the limit:
r_{2} - r_{1} Approaches 0, but is some infinitesimally small difference dr, which is explicitly non-zero.
The summation component:
r_{2} + r_{1} Seems to approach 2.

As I said before, writing r_2= r_1+ dr where dr is an "imfinitesmal" (a very technical term which I did not use before but use here because you did) we have (r_2- r_1)^2= r_1^2+ 2r_1dr+ (dr)^2- r_1^2= 2r_1dr+ (dr)^2= 2r_1dr+ (dr)^2= 2r_1dr because, while an infinitesmal is not 0, its square can treated as 0 in comparison with an infinitesmal.

So we have:
\lim \pi (r_{2} + r_{1})(r_{2} - r_{1}) = \pi(2r_{2}}{dr}

Is it considered to equal 2 or is it explicitly not 2 as the difference component is explicitly not 0?
I am confused, is there an error in my reasoning?
Any help appreciated as this is very frustrating!
Thanks in advance,
Nobahar

 

[nobahar]

Thanks for the response Halls.
Does my understanding of infintesimals appear accurate, even if not complete (I have often wondered on certain uses of infintesimals)? I hope if I can pursue differentials in the future my understanding will be improved; but does it at least appear to be 'along the correct lines'? I ask as you're reference to my use of infinitesimals seemed critical.
Many thanks, Nobahar.