Why does $int_0^R 2 pi r ,mathrm d r$ give the area of a circle?
$begingroup$
There's a method of computing the area of a circle by dividing it in concentric rings with infinitesimal width. Let $R$ be the radius of the circle and $r$ be the radius of the rings. The area of the circle is
$$int_0^R 2 pi r ,mathrm d r$$
My questions are:
I do not understand, though, how to justify the $2 pi r ,mathrm d r$ approximation for the area of each ring. Its actual area would be $$pi (r + mathrm d r)^2 - pi r^2 = 2 pi r ,mathrm d r + pi left( mathrm d r right)^2$$ right? Could I use this more precise formula if I wanted to? How?
The area on the integral above looks more like the lateral area of a cylinder of height $mathrm d r$, which is different from the actual area between two concentric circles. So why does that work?
Would the lateral area of a truncated cone (which seems to be the intermediate between the ring area and the cylinder lateral area) also work as an approximation?
Also, how do you come up with such an idea for an approximation that makes the calculation so beautifully simple (i.e. adopting the lateral area of a cylinder as the area of the rings)? It is considered a trivial integral, but there is a huge and mostly ignored step to be taken there.
calculus integration definite-integrals area polar-coordinates
edited Mar 18 at 9:18
user21820
39.8k544158
asked Mar 18 at 5:50
Jim KwonJim Kwon
411
$endgroup$
|
show 1 more comment
$begingroup$
There's a method of computing the area of a circle by dividing it in concentric rings with infinitesimal width. Let $R$ be the radius of the circle and $r$ be the radius of the rings. The area of the circle is
$$int_0^R 2 pi r ,mathrm d r$$
My questions are:
I do not understand, though, how to justify the $2 pi r ,mathrm d r$ approximation for the area of each ring. Its actual area would be $$pi (r + mathrm d r)^2 - pi r^2 = 2 pi r ,mathrm d r + pi left( mathrm d r right)^2$$ right? Could I use this more precise formula if I wanted to? How?
The area on the integral above looks more like the lateral area of a cylinder of height $mathrm d r$, which is different from the actual area between two concentric circles. So why does that work?
Would the lateral area of a truncated cone (which seems to be the intermediate between the ring area and the cylinder lateral area) also work as an approximation?
Also, how do you come up with such an idea for an approximation that makes the calculation so beautifully simple (i.e. adopting the lateral area of a cylinder as the area of the rings)? It is considered a trivial integral, but there is a huge and mostly ignored step to be taken there.
calculus integration definite-integrals area polar-coordinates
edited Mar 18 at 9:18
user21820
39.8k544158
asked Mar 18 at 5:50
Jim KwonJim Kwon
411
$endgroup$
6
$begingroup$
$pi(r + dr)^2 - pi r^2 = pi (r^2 + 2 r dr + (dr)^2- r^2) = pi(2 r dr + (dr)^2)$. As $dr$ goes to zero, $(dr)^2$ will approach zero more quickly than $dr$, so the expression is approximately equal to $2 pi r dr$, which is what you got in your integral.
$endgroup$
– 1123581321
Mar 18 at 5:51
$begingroup$
You are correct in your concern, and unless you want to get into differential forms this is best justified using Riemann sums. I wouldn't worry about it though - so long as you play with sufficiently nice shapes and spaces, differentials will work as you expect.
$endgroup$
– Brevan Ellefsen
Mar 18 at 5:57
$begingroup$
@1123581321 Will $(mathrm d r)^2$ approach zero more quickly, or is $(mathrm d r)^2 = 0$ by definition?
$endgroup$
– Rodrigo de Azevedo
Mar 18 at 6:41
$begingroup$
It will approach it more quickly since it is much less than $dr$ e.g. if $dr = 10^{-4}$, then $(dr)^2 = 10^{-8}$
$endgroup$
– 1123581321
Mar 18 at 6:42
4
$begingroup$
@RodrigodeAzevedo Whether $(dr)^2=0$ depends on how you define $dr$. If you're thinking of it as a small real change in a real quantity, the answer will be no (because in reals $x^2=0iff x=0$), but then I'd recommend calling it $delta r$ instead. If on the other hand you introduce so-called nilpotent infinitesimal $dr$, you can take $dr^2=0$ as a calculus axiom. (However, if you take that to heart beware the exact is-zero axioms depend on what you want to do in calculus, e.g. $ds^2=dx^2+dy^2$ wouldn't use that.)
$endgroup$
– J.G.
Mar 18 at 9:26
|
show 1 more comment
$begingroup$
There's a method of computing the area of a circle by dividing it in concentric rings with infinitesimal width. Let $R$ be the radius of the circle and $r$ be the radius of the rings. The area of the circle is
$$int_0^R 2 pi r ,mathrm d r$$
My questions are:
I do not understand, though, how to justify the $2 pi r ,mathrm d r$ approximation for the area of each ring. Its actual area would be $$pi (r + mathrm d r)^2 - pi r^2 = 2 pi r ,mathrm d r + pi left( mathrm d r right)^2$$ right? Could I use this more precise formula if I wanted to? How?
The area on the integral above looks more like the lateral area of a cylinder of height $mathrm d r$, which is different from the actual area between two concentric circles. So why does that work?
Would the lateral area of a truncated cone (which seems to be the intermediate between the ring area and the cylinder lateral area) also work as an approximation?
Also, how do you come up with such an idea for an approximation that makes the calculation so beautifully simple (i.e. adopting the lateral area of a cylinder as the area of the rings)? It is considered a trivial integral, but there is a huge and mostly ignored step to be taken there.
calculus integration definite-integrals area polar-coordinates
edited Mar 18 at 9:18
user21820
39.8k544158
asked Mar 18 at 5:50
Jim KwonJim Kwon
411
$endgroup$
There's a method of computing the area of a circle by dividing it in concentric rings with infinitesimal width. Let $R$ be the radius of the circle and $r$ be the radius of the rings. The area of the circle is
$$int_0^R 2 pi r ,mathrm d r$$
My questions are:
I do not understand, though, how to justify the $2 pi r ,mathrm d r$ approximation for the area of each ring. Its actual area would be $$pi (r + mathrm d r)^2 - pi r^2 = 2 pi r ,mathrm d r + pi left( mathrm d r right)^2$$ right? Could I use this more precise formula if I wanted to? How?
The area on the integral above looks more like the lateral area of a cylinder of height $mathrm d r$, which is different from the actual area between two concentric circles. So why does that work?
Would the lateral area of a truncated cone (which seems to be the intermediate between the ring area and the cylinder lateral area) also work as an approximation?
Also, how do you come up with such an idea for an approximation that makes the calculation so beautifully simple (i.e. adopting the lateral area of a cylinder as the area of the rings)? It is considered a trivial integral, but there is a huge and mostly ignored step to be taken there.
calculus integration definite-integrals area polar-coordinates
calculus integration definite-integrals area polar-coordinates
edited Mar 18 at 9:18
user21820
39.8k544158
asked Mar 18 at 5:50
Jim KwonJim Kwon
411
edited Mar 18 at 9:18
user21820
39.8k544158
asked Mar 18 at 5:50
Jim KwonJim Kwon
411
edited Mar 18 at 9:18
user21820
39.8k544158
edited Mar 18 at 9:18
user21820
39.8k544158
edited Mar 18 at 9:18
user21820
39.8k544158
39.8k544158
asked Mar 18 at 5:50
Jim KwonJim Kwon
411
asked Mar 18 at 5:50
Jim KwonJim Kwon
411
asked Mar 18 at 5:50
Jim KwonJim Kwon
411
411
6
$begingroup$
$pi(r + dr)^2 - pi r^2 = pi (r^2 + 2 r dr + (dr)^2- r^2) = pi(2 r dr + (dr)^2)$. As $dr$ goes to zero, $(dr)^2$ will approach zero more quickly than $dr$, so the expression is approximately equal to $2 pi r dr$, which is what you got in your integral.
$endgroup$
– 1123581321
Mar 18 at 5:51
$begingroup$
You are correct in your concern, and unless you want to get into differential forms this is best justified using Riemann sums. I wouldn't worry about it though - so long as you play with sufficiently nice shapes and spaces, differentials will work as you expect.
$endgroup$
– Brevan Ellefsen
Mar 18 at 5:57
$begingroup$
@1123581321 Will $(mathrm d r)^2$ approach zero more quickly, or is $(mathrm d r)^2 = 0$ by definition?
$endgroup$
– Rodrigo de Azevedo
Mar 18 at 6:41
$begingroup$
It will approach it more quickly since it is much less than $dr$ e.g. if $dr = 10^{-4}$, then $(dr)^2 = 10^{-8}$
$endgroup$
– 1123581321
Mar 18 at 6:42
4
$begingroup$
@RodrigodeAzevedo Whether $(dr)^2=0$ depends on how you define $dr$. If you're thinking of it as a small real change in a real quantity, the answer will be no (because in reals $x^2=0iff x=0$), but then I'd recommend calling it $delta r$ instead. If on the other hand you introduce so-called nilpotent infinitesimal $dr$, you can take $dr^2=0$ as a calculus axiom. (However, if you take that to heart beware the exact is-zero axioms depend on what you want to do in calculus, e.g. $ds^2=dx^2+dy^2$ wouldn't use that.)
$endgroup$
– J.G.
Mar 18 at 9:26
|
show 1 more comment
6
$begingroup$
$pi(r + dr)^2 - pi r^2 = pi (r^2 + 2 r dr + (dr)^2- r^2) = pi(2 r dr + (dr)^2)$. As $dr$ goes to zero, $(dr)^2$ will approach zero more quickly than $dr$, so the expression is approximately equal to $2 pi r dr$, which is what you got in your integral.
$endgroup$
– 1123581321
Mar 18 at 5:51
$begingroup$
You are correct in your concern, and unless you want to get into differential forms this is best justified using Riemann sums. I wouldn't worry about it though - so long as you play with sufficiently nice shapes and spaces, differentials will work as you expect.
$endgroup$
– Brevan Ellefsen
Mar 18 at 5:57
$begingroup$
@1123581321 Will $(mathrm d r)^2$ approach zero more quickly, or is $(mathrm d r)^2 = 0$ by definition?
$endgroup$
– Rodrigo de Azevedo
Mar 18 at 6:41
$begingroup$
It will approach it more quickly since it is much less than $dr$ e.g. if $dr = 10^{-4}$, then $(dr)^2 = 10^{-8}$
$endgroup$
– 1123581321
Mar 18 at 6:42
4
$begingroup$
@RodrigodeAzevedo Whether $(dr)^2=0$ depends on how you define $dr$. If you're thinking of it as a small real change in a real quantity, the answer will be no (because in reals $x^2=0iff x=0$), but then I'd recommend calling it $delta r$ instead. If on the other hand you introduce so-called nilpotent infinitesimal $dr$, you can take $dr^2=0$ as a calculus axiom. (However, if you take that to heart beware the exact is-zero axioms depend on what you want to do in calculus, e.g. $ds^2=dx^2+dy^2$ wouldn't use that.)
$endgroup$
– J.G.
Mar 18 at 9:26
6
6
$begingroup$
$pi(r + dr)^2 - pi r^2 = pi (r^2 + 2 r dr + (dr)^2- r^2) = pi(2 r dr + (dr)^2)$. As $dr$ goes to zero, $(dr)^2$ will approach zero more quickly than $dr$, so the expression is approximately equal to $2 pi r dr$, which is what you got in your integral.
$endgroup$
– 1123581321
Mar 18 at 5:51
$begingroup$
$pi(r + dr)^2 - pi r^2 = pi (r^2 + 2 r dr + (dr)^2- r^2) = pi(2 r dr + (dr)^2)$. As $dr$ goes to zero, $(dr)^2$ will approach zero more quickly than $dr$, so the expression is approximately equal to $2 pi r dr$, which is what you got in your integral.
$endgroup$
– 1123581321
Mar 18 at 5:51
$begingroup$
You are correct in your concern, and unless you want to get into differential forms this is best justified using Riemann sums. I wouldn't worry about it though - so long as you play with sufficiently nice shapes and spaces, differentials will work as you expect.
$endgroup$
– Brevan Ellefsen
Mar 18 at 5:57
$begingroup$
You are correct in your concern, and unless you want to get into differential forms this is best justified using Riemann sums. I wouldn't worry about it though - so long as you play with sufficiently nice shapes and spaces, differentials will work as you expect.
$endgroup$
– Brevan Ellefsen
Mar 18 at 5:57
$begingroup$
@1123581321 Will $(mathrm d r)^2$ approach zero more quickly, or is $(mathrm d r)^2 = 0$ by definition?
$endgroup$
– Rodrigo de Azevedo
Mar 18 at 6:41
$begingroup$
@1123581321 Will $(mathrm d r)^2$ approach zero more quickly, or is $(mathrm d r)^2 = 0$ by definition?
$endgroup$
– Rodrigo de Azevedo
Mar 18 at 6:41
$begingroup$
It will approach it more quickly since it is much less than $dr$ e.g. if $dr = 10^{-4}$, then $(dr)^2 = 10^{-8}$
$endgroup$
– 1123581321
Mar 18 at 6:42
$begingroup$
It will approach it more quickly since it is much less than $dr$ e.g. if $dr = 10^{-4}$, then $(dr)^2 = 10^{-8}$
$endgroup$
– 1123581321
Mar 18 at 6:42
4
4
$begingroup$
@RodrigodeAzevedo Whether $(dr)^2=0$ depends on how you define $dr$. If you're thinking of it as a small real change in a real quantity, the answer will be no (because in reals $x^2=0iff x=0$), but then I'd recommend calling it $delta r$ instead. If on the other hand you introduce so-called nilpotent infinitesimal $dr$, you can take $dr^2=0$ as a calculus axiom. (However, if you take that to heart beware the exact is-zero axioms depend on what you want to do in calculus, e.g. $ds^2=dx^2+dy^2$ wouldn't use that.)
$endgroup$
– J.G.
Mar 18 at 9:26
$begingroup$
@RodrigodeAzevedo Whether $(dr)^2=0$ depends on how you define $dr$. If you're thinking of it as a small real change in a real quantity, the answer will be no (because in reals $x^2=0iff x=0$), but then I'd recommend calling it $delta r$ instead. If on the other hand you introduce so-called nilpotent infinitesimal $dr$, you can take $dr^2=0$ as a calculus axiom. (However, if you take that to heart beware the exact is-zero axioms depend on what you want to do in calculus, e.g. $ds^2=dx^2+dy^2$ wouldn't use that.)
$endgroup$
– J.G.
Mar 18 at 9:26
|
show 1 more comment
1 Answer
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$begingroup$
The idea is that the approximation gets better and better as you make the rings finer and finer, so that you eventually end up with the approximate-style result even if you use the more exact version.
That the approximation is justified becomes clear if you think in terms of differentials instead, so that the width of the rings is just a difference; because as $Delta rto 0,$ the quadratic term in $Delta r$ approaches $0$ more quickly, or, as we say, is of smaller order than $Delta r.$ This is simply because $x^2=o(x),$ in general, as $xto 0.$
To answer the other questions, which you've numbered and reorganized:
The key idea to have in mind is that any initial approximation will do, provided the error vanishes in the limit sufficiently fast.
Interpretation in mathematics is multidimensional. The same expression can be interpreted differently depending on context or ease. Thus, for example, while Euclid would have thought of $xy$ (both factors positive reals) as a rectangular area, Descartes simply thought of it as a line segment. There are infinitely many other ways of thinking of this product; e.g., one group would be $underbrace{1×1×1×cdots×1}_{n} × xy,$ where $n$ is a positive integer.
Thus, you could well think of $2πrDelta r$ as the curved area of a cylinder. Unrolling this gives a rectangular strip of dimensions $2πr × Delta r.$ The annulus with median radius $r$ and width $Delta r$ can also be thought of as being represented (approximately) by that expression. Now as $Delta rto 0,$ the strip approaches a string of length $2πr,$ which is the same as the circumference of the circle that the annulus approaches too. The surface area of a conical frustum should work similarly.
answered Mar 18 at 6:23
AllawonderAllawonder
1
$endgroup$
add a comment |
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$begingroup$
The idea is that the approximation gets better and better as you make the rings finer and finer, so that you eventually end up with the approximate-style result even if you use the more exact version.
That the approximation is justified becomes clear if you think in terms of differentials instead, so that the width of the rings is just a difference; because as $Delta rto 0,$ the quadratic term in $Delta r$ approaches $0$ more quickly, or, as we say, is of smaller order than $Delta r.$ This is simply because $x^2=o(x),$ in general, as $xto 0.$
To answer the other questions, which you've numbered and reorganized:
The key idea to have in mind is that any initial approximation will do, provided the error vanishes in the limit sufficiently fast.
Interpretation in mathematics is multidimensional. The same expression can be interpreted differently depending on context or ease. Thus, for example, while Euclid would have thought of $xy$ (both factors positive reals) as a rectangular area, Descartes simply thought of it as a line segment. There are infinitely many other ways of thinking of this product; e.g., one group would be $underbrace{1×1×1×cdots×1}_{n} × xy,$ where $n$ is a positive integer.
Thus, you could well think of $2πrDelta r$ as the curved area of a cylinder. Unrolling this gives a rectangular strip of dimensions $2πr × Delta r.$ The annulus with median radius $r$ and width $Delta r$ can also be thought of as being represented (approximately) by that expression. Now as $Delta rto 0,$ the strip approaches a string of length $2πr,$ which is the same as the circumference of the circle that the annulus approaches too. The surface area of a conical frustum should work similarly.
answered Mar 18 at 6:23
AllawonderAllawonder
1
$endgroup$
add a comment |
$begingroup$
The idea is that the approximation gets better and better as you make the rings finer and finer, so that you eventually end up with the approximate-style result even if you use the more exact version.
That the approximation is justified becomes clear if you think in terms of differentials instead, so that the width of the rings is just a difference; because as $Delta rto 0,$ the quadratic term in $Delta r$ approaches $0$ more quickly, or, as we say, is of smaller order than $Delta r.$ This is simply because $x^2=o(x),$ in general, as $xto 0.$
To answer the other questions, which you've numbered and reorganized:
The key idea to have in mind is that any initial approximation will do, provided the error vanishes in the limit sufficiently fast.
Interpretation in mathematics is multidimensional. The same expression can be interpreted differently depending on context or ease. Thus, for example, while Euclid would have thought of $xy$ (both factors positive reals) as a rectangular area, Descartes simply thought of it as a line segment. There are infinitely many other ways of thinking of this product; e.g., one group would be $underbrace{1×1×1×cdots×1}_{n} × xy,$ where $n$ is a positive integer.
Thus, you could well think of $2πrDelta r$ as the curved area of a cylinder. Unrolling this gives a rectangular strip of dimensions $2πr × Delta r.$ The annulus with median radius $r$ and width $Delta r$ can also be thought of as being represented (approximately) by that expression. Now as $Delta rto 0,$ the strip approaches a string of length $2πr,$ which is the same as the circumference of the circle that the annulus approaches too. The surface area of a conical frustum should work similarly.
answered Mar 18 at 6:23
AllawonderAllawonder
1
$endgroup$
add a comment |
$begingroup$
The idea is that the approximation gets better and better as you make the rings finer and finer, so that you eventually end up with the approximate-style result even if you use the more exact version.
That the approximation is justified becomes clear if you think in terms of differentials instead, so that the width of the rings is just a difference; because as $Delta rto 0,$ the quadratic term in $Delta r$ approaches $0$ more quickly, or, as we say, is of smaller order than $Delta r.$ This is simply because $x^2=o(x),$ in general, as $xto 0.$
To answer the other questions, which you've numbered and reorganized:
The key idea to have in mind is that any initial approximation will do, provided the error vanishes in the limit sufficiently fast.
Interpretation in mathematics is multidimensional. The same expression can be interpreted differently depending on context or ease. Thus, for example, while Euclid would have thought of $xy$ (both factors positive reals) as a rectangular area, Descartes simply thought of it as a line segment. There are infinitely many other ways of thinking of this product; e.g., one group would be $underbrace{1×1×1×cdots×1}_{n} × xy,$ where $n$ is a positive integer.
Thus, you could well think of $2πrDelta r$ as the curved area of a cylinder. Unrolling this gives a rectangular strip of dimensions $2πr × Delta r.$ The annulus with median radius $r$ and width $Delta r$ can also be thought of as being represented (approximately) by that expression. Now as $Delta rto 0,$ the strip approaches a string of length $2πr,$ which is the same as the circumference of the circle that the annulus approaches too. The surface area of a conical frustum should work similarly.
answered Mar 18 at 6:23
AllawonderAllawonder
1
$endgroup$
The idea is that the approximation gets better and better as you make the rings finer and finer, so that you eventually end up with the approximate-style result even if you use the more exact version.
That the approximation is justified becomes clear if you think in terms of differentials instead, so that the width of the rings is just a difference; because as $Delta rto 0,$ the quadratic term in $Delta r$ approaches $0$ more quickly, or, as we say, is of smaller order than $Delta r.$ This is simply because $x^2=o(x),$ in general, as $xto 0.$
To answer the other questions, which you've numbered and reorganized:
The key idea to have in mind is that any initial approximation will do, provided the error vanishes in the limit sufficiently fast.
Interpretation in mathematics is multidimensional. The same expression can be interpreted differently depending on context or ease. Thus, for example, while Euclid would have thought of $xy$ (both factors positive reals) as a rectangular area, Descartes simply thought of it as a line segment. There are infinitely many other ways of thinking of this product; e.g., one group would be $underbrace{1×1×1×cdots×1}_{n} × xy,$ where $n$ is a positive integer.
Thus, you could well think of $2πrDelta r$ as the curved area of a cylinder. Unrolling this gives a rectangular strip of dimensions $2πr × Delta r.$ The annulus with median radius $r$ and width $Delta r$ can also be thought of as being represented (approximately) by that expression. Now as $Delta rto 0,$ the strip approaches a string of length $2πr,$ which is the same as the circumference of the circle that the annulus approaches too. The surface area of a conical frustum should work similarly.
answered Mar 18 at 6:23
AllawonderAllawonder
1
edited Mar 18 at 9:58
answered Mar 18 at 6:23
AllawonderAllawonder
1
answered Mar 18 at 6:23
AllawonderAllawonder
1
answered Mar 18 at 6:23
AllawonderAllawonder
1
1
add a comment |
add a comment |
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$begingroup$
$pi(r + dr)^2 - pi r^2 = pi (r^2 + 2 r dr + (dr)^2- r^2) = pi(2 r dr + (dr)^2)$. As $dr$ goes to zero, $(dr)^2$ will approach zero more quickly than $dr$, so the expression is approximately equal to $2 pi r dr$, which is what you got in your integral.
$endgroup$
– 1123581321
Mar 18 at 5:51
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You are correct in your concern, and unless you want to get into differential forms this is best justified using Riemann sums. I wouldn't worry about it though - so long as you play with sufficiently nice shapes and spaces, differentials will work as you expect.
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– Brevan Ellefsen
Mar 18 at 5:57
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@1123581321 Will $(mathrm d r)^2$ approach zero more quickly, or is $(mathrm d r)^2 = 0$ by definition?
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– Rodrigo de Azevedo
Mar 18 at 6:41
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It will approach it more quickly since it is much less than $dr$ e.g. if $dr = 10^{-4}$, then $(dr)^2 = 10^{-8}$
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– 1123581321
Mar 18 at 6:42
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@RodrigodeAzevedo Whether $(dr)^2=0$ depends on how you define $dr$. If you're thinking of it as a small real change in a real quantity, the answer will be no (because in reals $x^2=0iff x=0$), but then I'd recommend calling it $delta r$ instead. If on the other hand you introduce so-called nilpotent infinitesimal $dr$, you can take $dr^2=0$ as a calculus axiom. (However, if you take that to heart beware the exact is-zero axioms depend on what you want to do in calculus, e.g. $ds^2=dx^2+dy^2$ wouldn't use that.)
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– J.G.
Mar 18 at 9:26