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Mean value theorem for surface integrals: Shrinking surface of integration to $0$
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Consider a disk of radius $R$, area $A$, somewhere in $mathbbR^3$. Consider a vector function $vecF: mathbbR^3 to mathbbR^3$. I form the surface integral over the disk
$$ iint_textdisk vecF cdot dvecS \
iint_textdisk F_ndS
$$
where $F_n$ indicates the normal component to the oriented surface. I applied the Mean Value Theorem for integrals (which I'm assuming works for any integral of any dimension) to get
$$ F_n(P^*)A$$
where $P^*$ is some point on the disk. I form the equation
$$ frac1Aiint F_ndS = F_n(P^*)$$
Now I take the limit
$$ lim_Ato 0frac1Aiint F_ndS = lim_Ato0F_n(P^*) $$
What value will $F_n(P^*)$ converge to? Will it converge to the value over the center of the disk $F_n(P_0)$? I've done limits along paths in $mathbbR^2$ or $mathbbR^3$. However, I've never done limits of surface areas embedded in $mathbbR^3$. Is such an operation well defined? If the disk shrinks uniformly, I can imagine $F_n(P^*) to F_n(P_0)$ where $P_0$ is the center of the disk. However, if one side shrinks faster than the other, I can imagine $F_n(P^*)$ converging to any value we like depending on how we shrink the area. I could ask the same question for $1$-dimensional domains. Taking the limit after applying the Mean Value Theorem
$$ lim_hto0 frac1hint_a^a+h f(x) dx = lim_hto0 f(c) = f(a)$$
However, if I ask
$$ lim_ell to 0 frac1ell int_a^b f(x) dx$$
where $ell = b- a$ is the interval length, is such an operation well defined?
calculus integration limits multivariable-calculus
asked Jul 23 at 15:13
DWade64
3841313
 |Â
show 2 more comments
up vote
1
down vote
favorite
Consider a disk of radius $R$, area $A$, somewhere in $mathbbR^3$. Consider a vector function $vecF: mathbbR^3 to mathbbR^3$. I form the surface integral over the disk
$$ iint_textdisk vecF cdot dvecS \
iint_textdisk F_ndS
$$
where $F_n$ indicates the normal component to the oriented surface. I applied the Mean Value Theorem for integrals (which I'm assuming works for any integral of any dimension) to get
$$ F_n(P^*)A$$
where $P^*$ is some point on the disk. I form the equation
$$ frac1Aiint F_ndS = F_n(P^*)$$
Now I take the limit
$$ lim_Ato 0frac1Aiint F_ndS = lim_Ato0F_n(P^*) $$
What value will $F_n(P^*)$ converge to? Will it converge to the value over the center of the disk $F_n(P_0)$? I've done limits along paths in $mathbbR^2$ or $mathbbR^3$. However, I've never done limits of surface areas embedded in $mathbbR^3$. Is such an operation well defined? If the disk shrinks uniformly, I can imagine $F_n(P^*) to F_n(P_0)$ where $P_0$ is the center of the disk. However, if one side shrinks faster than the other, I can imagine $F_n(P^*)$ converging to any value we like depending on how we shrink the area. I could ask the same question for $1$-dimensional domains. Taking the limit after applying the Mean Value Theorem
$$ lim_hto0 frac1hint_a^a+h f(x) dx = lim_hto0 f(c) = f(a)$$
However, if I ask
$$ lim_ell to 0 frac1ell int_a^b f(x) dx$$
where $ell = b- a$ is the interval length, is such an operation well defined?
calculus integration limits multivariable-calculus
asked Jul 23 at 15:13
DWade64
3841313
The mean value theorem is in fact quite specific to one dimensional settings. In general, with several variables we get an inequality.
– Fimpellizieri
Jul 23 at 15:18
@Fimpellizieri I remember seeing a mean value theorem for double integrals in my textbook. Are you talking about the mean value theorem for derivatives?
– DWade64
Jul 23 at 15:21
We still don't get what you want with vector valued functions. The article linked has an example; see here.
– Fimpellizieri
Jul 23 at 15:27
@Fimpellizieri Ah that's interesting. Are there a class of functions which do satisfy the theorem? Because a surface integral is not much different than a double integral. It's still a volume over the surface where the height is given by $F_n$. It seems like it should work for some functions, but maybe not
– DWade64
Jul 23 at 15:50
Going from a single variable to multiple variables is a big change. I guess the most common but still powerful example is that things like saddles don't exist in one dimension.
– Fimpellizieri
Jul 23 at 15:57
 |Â
show 2 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Consider a disk of radius $R$, area $A$, somewhere in $mathbbR^3$. Consider a vector function $vecF: mathbbR^3 to mathbbR^3$. I form the surface integral over the disk
$$ iint_textdisk vecF cdot dvecS \
iint_textdisk F_ndS
$$
where $F_n$ indicates the normal component to the oriented surface. I applied the Mean Value Theorem for integrals (which I'm assuming works for any integral of any dimension) to get
$$ F_n(P^*)A$$
where $P^*$ is some point on the disk. I form the equation
$$ frac1Aiint F_ndS = F_n(P^*)$$
Now I take the limit
$$ lim_Ato 0frac1Aiint F_ndS = lim_Ato0F_n(P^*) $$
What value will $F_n(P^*)$ converge to? Will it converge to the value over the center of the disk $F_n(P_0)$? I've done limits along paths in $mathbbR^2$ or $mathbbR^3$. However, I've never done limits of surface areas embedded in $mathbbR^3$. Is such an operation well defined? If the disk shrinks uniformly, I can imagine $F_n(P^*) to F_n(P_0)$ where $P_0$ is the center of the disk. However, if one side shrinks faster than the other, I can imagine $F_n(P^*)$ converging to any value we like depending on how we shrink the area. I could ask the same question for $1$-dimensional domains. Taking the limit after applying the Mean Value Theorem
$$ lim_hto0 frac1hint_a^a+h f(x) dx = lim_hto0 f(c) = f(a)$$
However, if I ask
$$ lim_ell to 0 frac1ell int_a^b f(x) dx$$
where $ell = b- a$ is the interval length, is such an operation well defined?
calculus integration limits multivariable-calculus
asked Jul 23 at 15:13
DWade64
3841313
Consider a disk of radius $R$, area $A$, somewhere in $mathbbR^3$. Consider a vector function $vecF: mathbbR^3 to mathbbR^3$. I form the surface integral over the disk
$$ iint_textdisk vecF cdot dvecS \
iint_textdisk F_ndS
$$
where $F_n$ indicates the normal component to the oriented surface. I applied the Mean Value Theorem for integrals (which I'm assuming works for any integral of any dimension) to get
$$ F_n(P^*)A$$
where $P^*$ is some point on the disk. I form the equation
$$ frac1Aiint F_ndS = F_n(P^*)$$
Now I take the limit
$$ lim_Ato 0frac1Aiint F_ndS = lim_Ato0F_n(P^*) $$
What value will $F_n(P^*)$ converge to? Will it converge to the value over the center of the disk $F_n(P_0)$? I've done limits along paths in $mathbbR^2$ or $mathbbR^3$. However, I've never done limits of surface areas embedded in $mathbbR^3$. Is such an operation well defined? If the disk shrinks uniformly, I can imagine $F_n(P^*) to F_n(P_0)$ where $P_0$ is the center of the disk. However, if one side shrinks faster than the other, I can imagine $F_n(P^*)$ converging to any value we like depending on how we shrink the area. I could ask the same question for $1$-dimensional domains. Taking the limit after applying the Mean Value Theorem
$$ lim_hto0 frac1hint_a^a+h f(x) dx = lim_hto0 f(c) = f(a)$$
However, if I ask
$$ lim_ell to 0 frac1ell int_a^b f(x) dx$$
where $ell = b- a$ is the interval length, is such an operation well defined?
calculus integration limits multivariable-calculus
asked Jul 23 at 15:13
DWade64
3841313
edited Jul 23 at 15:23
asked Jul 23 at 15:13
DWade64
3841313
asked Jul 23 at 15:13
DWade64
3841313
asked Jul 23 at 15:13
DWade64
3841313
3841313
The mean value theorem is in fact quite specific to one dimensional settings. In general, with several variables we get an inequality.
– Fimpellizieri
Jul 23 at 15:18
@Fimpellizieri I remember seeing a mean value theorem for double integrals in my textbook. Are you talking about the mean value theorem for derivatives?
– DWade64
Jul 23 at 15:21
We still don't get what you want with vector valued functions. The article linked has an example; see here.
– Fimpellizieri
Jul 23 at 15:27
@Fimpellizieri Ah that's interesting. Are there a class of functions which do satisfy the theorem? Because a surface integral is not much different than a double integral. It's still a volume over the surface where the height is given by $F_n$. It seems like it should work for some functions, but maybe not
– DWade64
Jul 23 at 15:50
Going from a single variable to multiple variables is a big change. I guess the most common but still powerful example is that things like saddles don't exist in one dimension.
– Fimpellizieri
Jul 23 at 15:57
 |Â
show 2 more comments
The mean value theorem is in fact quite specific to one dimensional settings. In general, with several variables we get an inequality.
– Fimpellizieri
Jul 23 at 15:18
@Fimpellizieri I remember seeing a mean value theorem for double integrals in my textbook. Are you talking about the mean value theorem for derivatives?
– DWade64
Jul 23 at 15:21
We still don't get what you want with vector valued functions. The article linked has an example; see here.
– Fimpellizieri
Jul 23 at 15:27
@Fimpellizieri Ah that's interesting. Are there a class of functions which do satisfy the theorem? Because a surface integral is not much different than a double integral. It's still a volume over the surface where the height is given by $F_n$. It seems like it should work for some functions, but maybe not
– DWade64
Jul 23 at 15:50
Going from a single variable to multiple variables is a big change. I guess the most common but still powerful example is that things like saddles don't exist in one dimension.
– Fimpellizieri
Jul 23 at 15:57
The mean value theorem is in fact quite specific to one dimensional settings. In general, with several variables we get an inequality.
– Fimpellizieri
Jul 23 at 15:18
The mean value theorem is in fact quite specific to one dimensional settings. In general, with several variables we get an inequality.
– Fimpellizieri
Jul 23 at 15:18
@Fimpellizieri I remember seeing a mean value theorem for double integrals in my textbook. Are you talking about the mean value theorem for derivatives?
– DWade64
Jul 23 at 15:21
@Fimpellizieri I remember seeing a mean value theorem for double integrals in my textbook. Are you talking about the mean value theorem for derivatives?
– DWade64
Jul 23 at 15:21
We still don't get what you want with vector valued functions. The article linked has an example; see here.
– Fimpellizieri
Jul 23 at 15:27
We still don't get what you want with vector valued functions. The article linked has an example; see here.
– Fimpellizieri
Jul 23 at 15:27
@Fimpellizieri Ah that's interesting. Are there a class of functions which do satisfy the theorem? Because a surface integral is not much different than a double integral. It's still a volume over the surface where the height is given by $F_n$. It seems like it should work for some functions, but maybe not
– DWade64
Jul 23 at 15:50
@Fimpellizieri Ah that's interesting. Are there a class of functions which do satisfy the theorem? Because a surface integral is not much different than a double integral. It's still a volume over the surface where the height is given by $F_n$. It seems like it should work for some functions, but maybe not
– DWade64
Jul 23 at 15:50
Going from a single variable to multiple variables is a big change. I guess the most common but still powerful example is that things like saddles don't exist in one dimension.
– Fimpellizieri
Jul 23 at 15:57
Going from a single variable to multiple variables is a big change. I guess the most common but still powerful example is that things like saddles don't exist in one dimension.
– Fimpellizieri
Jul 23 at 15:57
 |Â
show 2 more comments
2 Answers
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up vote
1
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accepted
There is a mean value theorem for multiple integrals. For example, if $f:U subset mathbbR^2 to mathbbR$ is continuous and $U$ is compact and rectifiable, then there is a point $xi in U$, not necessarily unique, such that
$$int_U f = f(xi)cdot textarea(U)$$
This is proved in the usual way, noting that if $f$ attains minimum (maximum) values $m (M)$, then $m cdot textarea(U) leqslant int_Uf leqslant M cdot textarea(U).$
It seems you have in mind an integral over a "nice" surface that can be parameterized as $S = ,mathbfx = mathbfg(s,t), ,(s,t)in U subset mathbbR^2$, where $mathbfg in C^1(U)$ and $U$ is compact and rectifiable.
The surface integral can be expressed as
$$int_S F_n , dS= intint_U F_n(mathbfg(s,t)) left|fracpartial mathbfgpartial s times fracpartial mathbfgpartial t right|, ds,dt,$$
and if $F_n$ is continuous, the entire integrand is continuous. Whence, there is a point $(xi,eta) in U$ such that
$$int_S F_n , dS = F_n(mathbfg(xi,eta)) left|fracpartial mathbfgpartial s times fracpartial mathbfgpartial t right|_(xi,eta)cdottextarea(U)$$
Is "shrinking" the surface is a well-defined operation? The answer is only if you specify the operation as such. For example, construct a decreasing sequence of subsets $S_k$ where $S_k+1 subset S_k subset S$ and $bigcap_n=1^infty S_k = mathbfq $. There will be a corresponding sequence of subsets $U_k subset U$ and a point $(s_q,t_q) = mathbfg^-1(mathbfq)$.
Applying the mean value theorem we find a sequence $(xi_k,eta_k)$ converging to $(s_q,t_q)$ such that
$$frac1textarea(S_k) int_S_k F_n , dS = F_n(mathbfg(xi_k,eta_k)) left|fracpartial mathbfgpartial s times fracpartial mathbfgpartial t right|_(xi_k,eta_k) cdot fractextarea(U_k)textarea(S_k),$$
which should converge as $k to infty$ to $F_n(mathbfq)$.
To reiterate, the particular value of the limit will depend on how the shrinking process is specified, as illustrated by taking your example further:
$$lim_h to 0 frac1hint_a^a+h f(x) , dx = f(a), \ lim_h to 0 frac1hint_b-h^b f(x) , dx = f(b)$$
answered Jul 23 at 18:11
RRL
43.6k42260
Thank you for this really nice answer! Motivation for this question: consider a surface $S$ with oriented normal vectors. The surface has a charge density $sigma(S)$. As you go from one side of the surface to the over, the electric field undergoes a discontinuity in its normal component to the surface of size $sigma/epsilon_0$. Every source on the internet, and many textbooks never give a proof of this very very important result! They use Gauss's law $oint vecE cdot dvecS = int sigma dS$. Everyone says "let your Gaussian surface be a cylinder through a patch of charge"
– DWade64
Jul 23 at 19:02
They say 'let the patch be small so that $sigma$ is approximately constant.' Then they say let the length of the cylinder go to zero so you get no flux through the sides of the cylinder. Then because the patch is small (top's and bottom cylinder lids are small), this some how makes the E field constant and blah blah. They have 2 limits going on. The cylinder length and size of lids A. Somehow letting $A$ go to zero makes the normal vectors of the gaussian surface, equal to the normal vector of the surface of interest with the charge. And many other assumptions.Thanks again
– DWade64
Jul 23 at 19:07
@DWade64: You're welcome.
– RRL
Jul 23 at 19:53
I'm using a multivariable calculus textbook designed for a first pass through (author Jon Rogawski). If I may pose this question here (as opposed to starting a new question). Consider a map $G: Dsubset R^2 to R^2$. He writes Area(G(D)) $approx |J_G$|(P) Area(D), where D is small, $J_G$ is the jacobian matrix, and P is a sample point in D. He writes, the more precise relationship is |$J_G$|(P) = $lim_$ Area(G(D))/Area(D), where $|D|to 0$ indicates the limit as the diameter of D (maximum distance b/w 2 points in D) tends to 0. What point P does he have in mind? Is it well-defined?
– DWade64
Jul 24 at 13:56
1
That's a pretty vague argument, particularly since it uses the $approx$ notion which is not precise. It would be better to use Area(G(D)) = $int_D 1 cdot |J_G|$ and apply the mean value theorem. In the end it reduces to $|J_G|(P)$ so its clear he has a specific point $P$ in mind and the shrinking procedure is defined through a sequence of sets $D_n$ or a continuous transformation $t mapsto D_t$ where $P$ is always an element of those sets.
– RRL
Jul 25 at 2:58
 |Â
show 1 more comment
up vote
1
down vote
If the vector field $bf v$ is continuous in a neighborhood of $P_0$ then for any disc $D$ with center $P_0$ and unit normal $bf n$ one has
$$int_Dbf vcdotbf n>rm domega= bf v(P_*)cdotbf n rm area(D)$$
for some $P_*in D$. This follows from the standard MVT for continuous real-valued functions on a path-connected set $Dsubsetmathbb R^3$. It follows that
$$lim_r(D)to 0+1overrm area(D)int_Dbf vcdotbf n>rm domega= bf v(P_0)cdotbf n .$$
But you can also set up an $epsilon/delta$-proof that does not make use of the MVT.
answered Jul 23 at 19:10
Christian Blatter
163k7107306
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
There is a mean value theorem for multiple integrals. For example, if $f:U subset mathbbR^2 to mathbbR$ is continuous and $U$ is compact and rectifiable, then there is a point $xi in U$, not necessarily unique, such that
$$int_U f = f(xi)cdot textarea(U)$$
This is proved in the usual way, noting that if $f$ attains minimum (maximum) values $m (M)$, then $m cdot textarea(U) leqslant int_Uf leqslant M cdot textarea(U).$
It seems you have in mind an integral over a "nice" surface that can be parameterized as $S = ,mathbfx = mathbfg(s,t), ,(s,t)in U subset mathbbR^2$, where $mathbfg in C^1(U)$ and $U$ is compact and rectifiable.
The surface integral can be expressed as
$$int_S F_n , dS= intint_U F_n(mathbfg(s,t)) left|fracpartial mathbfgpartial s times fracpartial mathbfgpartial t right|, ds,dt,$$
and if $F_n$ is continuous, the entire integrand is continuous. Whence, there is a point $(xi,eta) in U$ such that
$$int_S F_n , dS = F_n(mathbfg(xi,eta)) left|fracpartial mathbfgpartial s times fracpartial mathbfgpartial t right|_(xi,eta)cdottextarea(U)$$
Is "shrinking" the surface is a well-defined operation? The answer is only if you specify the operation as such. For example, construct a decreasing sequence of subsets $S_k$ where $S_k+1 subset S_k subset S$ and $bigcap_n=1^infty S_k = mathbfq $. There will be a corresponding sequence of subsets $U_k subset U$ and a point $(s_q,t_q) = mathbfg^-1(mathbfq)$.
Applying the mean value theorem we find a sequence $(xi_k,eta_k)$ converging to $(s_q,t_q)$ such that
$$frac1textarea(S_k) int_S_k F_n , dS = F_n(mathbfg(xi_k,eta_k)) left|fracpartial mathbfgpartial s times fracpartial mathbfgpartial t right|_(xi_k,eta_k) cdot fractextarea(U_k)textarea(S_k),$$
which should converge as $k to infty$ to $F_n(mathbfq)$.
To reiterate, the particular value of the limit will depend on how the shrinking process is specified, as illustrated by taking your example further:
$$lim_h to 0 frac1hint_a^a+h f(x) , dx = f(a), \ lim_h to 0 frac1hint_b-h^b f(x) , dx = f(b)$$
answered Jul 23 at 18:11
RRL
43.6k42260
Thank you for this really nice answer! Motivation for this question: consider a surface $S$ with oriented normal vectors. The surface has a charge density $sigma(S)$. As you go from one side of the surface to the over, the electric field undergoes a discontinuity in its normal component to the surface of size $sigma/epsilon_0$. Every source on the internet, and many textbooks never give a proof of this very very important result! They use Gauss's law $oint vecE cdot dvecS = int sigma dS$. Everyone says "let your Gaussian surface be a cylinder through a patch of charge"
– DWade64
Jul 23 at 19:02
They say 'let the patch be small so that $sigma$ is approximately constant.' Then they say let the length of the cylinder go to zero so you get no flux through the sides of the cylinder. Then because the patch is small (top's and bottom cylinder lids are small), this some how makes the E field constant and blah blah. They have 2 limits going on. The cylinder length and size of lids A. Somehow letting $A$ go to zero makes the normal vectors of the gaussian surface, equal to the normal vector of the surface of interest with the charge. And many other assumptions.Thanks again
– DWade64
Jul 23 at 19:07
@DWade64: You're welcome.
– RRL
Jul 23 at 19:53
I'm using a multivariable calculus textbook designed for a first pass through (author Jon Rogawski). If I may pose this question here (as opposed to starting a new question). Consider a map $G: Dsubset R^2 to R^2$. He writes Area(G(D)) $approx |J_G$|(P) Area(D), where D is small, $J_G$ is the jacobian matrix, and P is a sample point in D. He writes, the more precise relationship is |$J_G$|(P) = $lim_$ Area(G(D))/Area(D), where $|D|to 0$ indicates the limit as the diameter of D (maximum distance b/w 2 points in D) tends to 0. What point P does he have in mind? Is it well-defined?
– DWade64
Jul 24 at 13:56
1
That's a pretty vague argument, particularly since it uses the $approx$ notion which is not precise. It would be better to use Area(G(D)) = $int_D 1 cdot |J_G|$ and apply the mean value theorem. In the end it reduces to $|J_G|(P)$ so its clear he has a specific point $P$ in mind and the shrinking procedure is defined through a sequence of sets $D_n$ or a continuous transformation $t mapsto D_t$ where $P$ is always an element of those sets.
– RRL
Jul 25 at 2:58
 |Â
show 1 more comment
up vote
1
down vote
accepted
There is a mean value theorem for multiple integrals. For example, if $f:U subset mathbbR^2 to mathbbR$ is continuous and $U$ is compact and rectifiable, then there is a point $xi in U$, not necessarily unique, such that
$$int_U f = f(xi)cdot textarea(U)$$
This is proved in the usual way, noting that if $f$ attains minimum (maximum) values $m (M)$, then $m cdot textarea(U) leqslant int_Uf leqslant M cdot textarea(U).$
It seems you have in mind an integral over a "nice" surface that can be parameterized as $S = ,mathbfx = mathbfg(s,t), ,(s,t)in U subset mathbbR^2$, where $mathbfg in C^1(U)$ and $U$ is compact and rectifiable.
The surface integral can be expressed as
$$int_S F_n , dS= intint_U F_n(mathbfg(s,t)) left|fracpartial mathbfgpartial s times fracpartial mathbfgpartial t right|, ds,dt,$$
and if $F_n$ is continuous, the entire integrand is continuous. Whence, there is a point $(xi,eta) in U$ such that
$$int_S F_n , dS = F_n(mathbfg(xi,eta)) left|fracpartial mathbfgpartial s times fracpartial mathbfgpartial t right|_(xi,eta)cdottextarea(U)$$
Is "shrinking" the surface is a well-defined operation? The answer is only if you specify the operation as such. For example, construct a decreasing sequence of subsets $S_k$ where $S_k+1 subset S_k subset S$ and $bigcap_n=1^infty S_k = mathbfq $. There will be a corresponding sequence of subsets $U_k subset U$ and a point $(s_q,t_q) = mathbfg^-1(mathbfq)$.
Applying the mean value theorem we find a sequence $(xi_k,eta_k)$ converging to $(s_q,t_q)$ such that
$$frac1textarea(S_k) int_S_k F_n , dS = F_n(mathbfg(xi_k,eta_k)) left|fracpartial mathbfgpartial s times fracpartial mathbfgpartial t right|_(xi_k,eta_k) cdot fractextarea(U_k)textarea(S_k),$$
which should converge as $k to infty$ to $F_n(mathbfq)$.
To reiterate, the particular value of the limit will depend on how the shrinking process is specified, as illustrated by taking your example further:
$$lim_h to 0 frac1hint_a^a+h f(x) , dx = f(a), \ lim_h to 0 frac1hint_b-h^b f(x) , dx = f(b)$$
answered Jul 23 at 18:11
RRL
43.6k42260
Thank you for this really nice answer! Motivation for this question: consider a surface $S$ with oriented normal vectors. The surface has a charge density $sigma(S)$. As you go from one side of the surface to the over, the electric field undergoes a discontinuity in its normal component to the surface of size $sigma/epsilon_0$. Every source on the internet, and many textbooks never give a proof of this very very important result! They use Gauss's law $oint vecE cdot dvecS = int sigma dS$. Everyone says "let your Gaussian surface be a cylinder through a patch of charge"
– DWade64
Jul 23 at 19:02
They say 'let the patch be small so that $sigma$ is approximately constant.' Then they say let the length of the cylinder go to zero so you get no flux through the sides of the cylinder. Then because the patch is small (top's and bottom cylinder lids are small), this some how makes the E field constant and blah blah. They have 2 limits going on. The cylinder length and size of lids A. Somehow letting $A$ go to zero makes the normal vectors of the gaussian surface, equal to the normal vector of the surface of interest with the charge. And many other assumptions.Thanks again
– DWade64
Jul 23 at 19:07
@DWade64: You're welcome.
– RRL
Jul 23 at 19:53
I'm using a multivariable calculus textbook designed for a first pass through (author Jon Rogawski). If I may pose this question here (as opposed to starting a new question). Consider a map $G: Dsubset R^2 to R^2$. He writes Area(G(D)) $approx |J_G$|(P) Area(D), where D is small, $J_G$ is the jacobian matrix, and P is a sample point in D. He writes, the more precise relationship is |$J_G$|(P) = $lim_$ Area(G(D))/Area(D), where $|D|to 0$ indicates the limit as the diameter of D (maximum distance b/w 2 points in D) tends to 0. What point P does he have in mind? Is it well-defined?
– DWade64
Jul 24 at 13:56
1
That's a pretty vague argument, particularly since it uses the $approx$ notion which is not precise. It would be better to use Area(G(D)) = $int_D 1 cdot |J_G|$ and apply the mean value theorem. In the end it reduces to $|J_G|(P)$ so its clear he has a specific point $P$ in mind and the shrinking procedure is defined through a sequence of sets $D_n$ or a continuous transformation $t mapsto D_t$ where $P$ is always an element of those sets.
– RRL
Jul 25 at 2:58
 |Â
show 1 more comment
up vote
1
down vote
accepted
up vote
1
down vote
accepted
There is a mean value theorem for multiple integrals. For example, if $f:U subset mathbbR^2 to mathbbR$ is continuous and $U$ is compact and rectifiable, then there is a point $xi in U$, not necessarily unique, such that
$$int_U f = f(xi)cdot textarea(U)$$
This is proved in the usual way, noting that if $f$ attains minimum (maximum) values $m (M)$, then $m cdot textarea(U) leqslant int_Uf leqslant M cdot textarea(U).$
It seems you have in mind an integral over a "nice" surface that can be parameterized as $S = ,mathbfx = mathbfg(s,t), ,(s,t)in U subset mathbbR^2$, where $mathbfg in C^1(U)$ and $U$ is compact and rectifiable.
The surface integral can be expressed as
$$int_S F_n , dS= intint_U F_n(mathbfg(s,t)) left|fracpartial mathbfgpartial s times fracpartial mathbfgpartial t right|, ds,dt,$$
and if $F_n$ is continuous, the entire integrand is continuous. Whence, there is a point $(xi,eta) in U$ such that
$$int_S F_n , dS = F_n(mathbfg(xi,eta)) left|fracpartial mathbfgpartial s times fracpartial mathbfgpartial t right|_(xi,eta)cdottextarea(U)$$
Is "shrinking" the surface is a well-defined operation? The answer is only if you specify the operation as such. For example, construct a decreasing sequence of subsets $S_k$ where $S_k+1 subset S_k subset S$ and $bigcap_n=1^infty S_k = mathbfq $. There will be a corresponding sequence of subsets $U_k subset U$ and a point $(s_q,t_q) = mathbfg^-1(mathbfq)$.
Applying the mean value theorem we find a sequence $(xi_k,eta_k)$ converging to $(s_q,t_q)$ such that
$$frac1textarea(S_k) int_S_k F_n , dS = F_n(mathbfg(xi_k,eta_k)) left|fracpartial mathbfgpartial s times fracpartial mathbfgpartial t right|_(xi_k,eta_k) cdot fractextarea(U_k)textarea(S_k),$$
which should converge as $k to infty$ to $F_n(mathbfq)$.
To reiterate, the particular value of the limit will depend on how the shrinking process is specified, as illustrated by taking your example further:
$$lim_h to 0 frac1hint_a^a+h f(x) , dx = f(a), \ lim_h to 0 frac1hint_b-h^b f(x) , dx = f(b)$$
answered Jul 23 at 18:11
RRL
43.6k42260
There is a mean value theorem for multiple integrals. For example, if $f:U subset mathbbR^2 to mathbbR$ is continuous and $U$ is compact and rectifiable, then there is a point $xi in U$, not necessarily unique, such that
$$int_U f = f(xi)cdot textarea(U)$$
This is proved in the usual way, noting that if $f$ attains minimum (maximum) values $m (M)$, then $m cdot textarea(U) leqslant int_Uf leqslant M cdot textarea(U).$
It seems you have in mind an integral over a "nice" surface that can be parameterized as $S = ,mathbfx = mathbfg(s,t), ,(s,t)in U subset mathbbR^2$, where $mathbfg in C^1(U)$ and $U$ is compact and rectifiable.
The surface integral can be expressed as
$$int_S F_n , dS= intint_U F_n(mathbfg(s,t)) left|fracpartial mathbfgpartial s times fracpartial mathbfgpartial t right|, ds,dt,$$
and if $F_n$ is continuous, the entire integrand is continuous. Whence, there is a point $(xi,eta) in U$ such that
$$int_S F_n , dS = F_n(mathbfg(xi,eta)) left|fracpartial mathbfgpartial s times fracpartial mathbfgpartial t right|_(xi,eta)cdottextarea(U)$$
Is "shrinking" the surface is a well-defined operation? The answer is only if you specify the operation as such. For example, construct a decreasing sequence of subsets $S_k$ where $S_k+1 subset S_k subset S$ and $bigcap_n=1^infty S_k = mathbfq $. There will be a corresponding sequence of subsets $U_k subset U$ and a point $(s_q,t_q) = mathbfg^-1(mathbfq)$.
Applying the mean value theorem we find a sequence $(xi_k,eta_k)$ converging to $(s_q,t_q)$ such that
$$frac1textarea(S_k) int_S_k F_n , dS = F_n(mathbfg(xi_k,eta_k)) left|fracpartial mathbfgpartial s times fracpartial mathbfgpartial t right|_(xi_k,eta_k) cdot fractextarea(U_k)textarea(S_k),$$
which should converge as $k to infty$ to $F_n(mathbfq)$.
To reiterate, the particular value of the limit will depend on how the shrinking process is specified, as illustrated by taking your example further:
$$lim_h to 0 frac1hint_a^a+h f(x) , dx = f(a), \ lim_h to 0 frac1hint_b-h^b f(x) , dx = f(b)$$
answered Jul 23 at 18:11
RRL
43.6k42260
answered Jul 23 at 18:11
RRL
43.6k42260
answered Jul 23 at 18:11
RRL
43.6k42260
answered Jul 23 at 18:11
RRL
43.6k42260
43.6k42260
Thank you for this really nice answer! Motivation for this question: consider a surface $S$ with oriented normal vectors. The surface has a charge density $sigma(S)$. As you go from one side of the surface to the over, the electric field undergoes a discontinuity in its normal component to the surface of size $sigma/epsilon_0$. Every source on the internet, and many textbooks never give a proof of this very very important result! They use Gauss's law $oint vecE cdot dvecS = int sigma dS$. Everyone says "let your Gaussian surface be a cylinder through a patch of charge"
– DWade64
Jul 23 at 19:02
They say 'let the patch be small so that $sigma$ is approximately constant.' Then they say let the length of the cylinder go to zero so you get no flux through the sides of the cylinder. Then because the patch is small (top's and bottom cylinder lids are small), this some how makes the E field constant and blah blah. They have 2 limits going on. The cylinder length and size of lids A. Somehow letting $A$ go to zero makes the normal vectors of the gaussian surface, equal to the normal vector of the surface of interest with the charge. And many other assumptions.Thanks again
– DWade64
Jul 23 at 19:07
@DWade64: You're welcome.
– RRL
Jul 23 at 19:53
I'm using a multivariable calculus textbook designed for a first pass through (author Jon Rogawski). If I may pose this question here (as opposed to starting a new question). Consider a map $G: Dsubset R^2 to R^2$. He writes Area(G(D)) $approx |J_G$|(P) Area(D), where D is small, $J_G$ is the jacobian matrix, and P is a sample point in D. He writes, the more precise relationship is |$J_G$|(P) = $lim_$ Area(G(D))/Area(D), where $|D|to 0$ indicates the limit as the diameter of D (maximum distance b/w 2 points in D) tends to 0. What point P does he have in mind? Is it well-defined?
– DWade64
Jul 24 at 13:56
1
That's a pretty vague argument, particularly since it uses the $approx$ notion which is not precise. It would be better to use Area(G(D)) = $int_D 1 cdot |J_G|$ and apply the mean value theorem. In the end it reduces to $|J_G|(P)$ so its clear he has a specific point $P$ in mind and the shrinking procedure is defined through a sequence of sets $D_n$ or a continuous transformation $t mapsto D_t$ where $P$ is always an element of those sets.
– RRL
Jul 25 at 2:58
 |Â
show 1 more comment
Thank you for this really nice answer! Motivation for this question: consider a surface $S$ with oriented normal vectors. The surface has a charge density $sigma(S)$. As you go from one side of the surface to the over, the electric field undergoes a discontinuity in its normal component to the surface of size $sigma/epsilon_0$. Every source on the internet, and many textbooks never give a proof of this very very important result! They use Gauss's law $oint vecE cdot dvecS = int sigma dS$. Everyone says "let your Gaussian surface be a cylinder through a patch of charge"
– DWade64
Jul 23 at 19:02
They say 'let the patch be small so that $sigma$ is approximately constant.' Then they say let the length of the cylinder go to zero so you get no flux through the sides of the cylinder. Then because the patch is small (top's and bottom cylinder lids are small), this some how makes the E field constant and blah blah. They have 2 limits going on. The cylinder length and size of lids A. Somehow letting $A$ go to zero makes the normal vectors of the gaussian surface, equal to the normal vector of the surface of interest with the charge. And many other assumptions.Thanks again
– DWade64
Jul 23 at 19:07
@DWade64: You're welcome.
– RRL
Jul 23 at 19:53
I'm using a multivariable calculus textbook designed for a first pass through (author Jon Rogawski). If I may pose this question here (as opposed to starting a new question). Consider a map $G: Dsubset R^2 to R^2$. He writes Area(G(D)) $approx |J_G$|(P) Area(D), where D is small, $J_G$ is the jacobian matrix, and P is a sample point in D. He writes, the more precise relationship is |$J_G$|(P) = $lim_$ Area(G(D))/Area(D), where $|D|to 0$ indicates the limit as the diameter of D (maximum distance b/w 2 points in D) tends to 0. What point P does he have in mind? Is it well-defined?
– DWade64
Jul 24 at 13:56
1
That's a pretty vague argument, particularly since it uses the $approx$ notion which is not precise. It would be better to use Area(G(D)) = $int_D 1 cdot |J_G|$ and apply the mean value theorem. In the end it reduces to $|J_G|(P)$ so its clear he has a specific point $P$ in mind and the shrinking procedure is defined through a sequence of sets $D_n$ or a continuous transformation $t mapsto D_t$ where $P$ is always an element of those sets.
– RRL
Jul 25 at 2:58
Thank you for this really nice answer! Motivation for this question: consider a surface $S$ with oriented normal vectors. The surface has a charge density $sigma(S)$. As you go from one side of the surface to the over, the electric field undergoes a discontinuity in its normal component to the surface of size $sigma/epsilon_0$. Every source on the internet, and many textbooks never give a proof of this very very important result! They use Gauss's law $oint vecE cdot dvecS = int sigma dS$. Everyone says "let your Gaussian surface be a cylinder through a patch of charge"
– DWade64
Jul 23 at 19:02
Thank you for this really nice answer! Motivation for this question: consider a surface $S$ with oriented normal vectors. The surface has a charge density $sigma(S)$. As you go from one side of the surface to the over, the electric field undergoes a discontinuity in its normal component to the surface of size $sigma/epsilon_0$. Every source on the internet, and many textbooks never give a proof of this very very important result! They use Gauss's law $oint vecE cdot dvecS = int sigma dS$. Everyone says "let your Gaussian surface be a cylinder through a patch of charge"
– DWade64
Jul 23 at 19:02
They say 'let the patch be small so that $sigma$ is approximately constant.' Then they say let the length of the cylinder go to zero so you get no flux through the sides of the cylinder. Then because the patch is small (top's and bottom cylinder lids are small), this some how makes the E field constant and blah blah. They have 2 limits going on. The cylinder length and size of lids A. Somehow letting $A$ go to zero makes the normal vectors of the gaussian surface, equal to the normal vector of the surface of interest with the charge. And many other assumptions.Thanks again
– DWade64
Jul 23 at 19:07
They say 'let the patch be small so that $sigma$ is approximately constant.' Then they say let the length of the cylinder go to zero so you get no flux through the sides of the cylinder. Then because the patch is small (top's and bottom cylinder lids are small), this some how makes the E field constant and blah blah. They have 2 limits going on. The cylinder length and size of lids A. Somehow letting $A$ go to zero makes the normal vectors of the gaussian surface, equal to the normal vector of the surface of interest with the charge. And many other assumptions.Thanks again
– DWade64
Jul 23 at 19:07
@DWade64: You're welcome.
– RRL
Jul 23 at 19:53
@DWade64: You're welcome.
– RRL
Jul 23 at 19:53
I'm using a multivariable calculus textbook designed for a first pass through (author Jon Rogawski). If I may pose this question here (as opposed to starting a new question). Consider a map $G: Dsubset R^2 to R^2$. He writes Area(G(D)) $approx |J_G$|(P) Area(D), where D is small, $J_G$ is the jacobian matrix, and P is a sample point in D. He writes, the more precise relationship is |$J_G$|(P) = $lim_$ Area(G(D))/Area(D), where $|D|to 0$ indicates the limit as the diameter of D (maximum distance b/w 2 points in D) tends to 0. What point P does he have in mind? Is it well-defined?
– DWade64
Jul 24 at 13:56
I'm using a multivariable calculus textbook designed for a first pass through (author Jon Rogawski). If I may pose this question here (as opposed to starting a new question). Consider a map $G: Dsubset R^2 to R^2$. He writes Area(G(D)) $approx |J_G$|(P) Area(D), where D is small, $J_G$ is the jacobian matrix, and P is a sample point in D. He writes, the more precise relationship is |$J_G$|(P) = $lim_$ Area(G(D))/Area(D), where $|D|to 0$ indicates the limit as the diameter of D (maximum distance b/w 2 points in D) tends to 0. What point P does he have in mind? Is it well-defined?
– DWade64
Jul 24 at 13:56
1
1
That's a pretty vague argument, particularly since it uses the $approx$ notion which is not precise. It would be better to use Area(G(D)) = $int_D 1 cdot |J_G|$ and apply the mean value theorem. In the end it reduces to $|J_G|(P)$ so its clear he has a specific point $P$ in mind and the shrinking procedure is defined through a sequence of sets $D_n$ or a continuous transformation $t mapsto D_t$ where $P$ is always an element of those sets.
– RRL
Jul 25 at 2:58
That's a pretty vague argument, particularly since it uses the $approx$ notion which is not precise. It would be better to use Area(G(D)) = $int_D 1 cdot |J_G|$ and apply the mean value theorem. In the end it reduces to $|J_G|(P)$ so its clear he has a specific point $P$ in mind and the shrinking procedure is defined through a sequence of sets $D_n$ or a continuous transformation $t mapsto D_t$ where $P$ is always an element of those sets.
– RRL
Jul 25 at 2:58
 |Â
show 1 more comment
up vote
1
down vote
If the vector field $bf v$ is continuous in a neighborhood of $P_0$ then for any disc $D$ with center $P_0$ and unit normal $bf n$ one has
$$int_Dbf vcdotbf n>rm domega= bf v(P_*)cdotbf n rm area(D)$$
for some $P_*in D$. This follows from the standard MVT for continuous real-valued functions on a path-connected set $Dsubsetmathbb R^3$. It follows that
$$lim_r(D)to 0+1overrm area(D)int_Dbf vcdotbf n>rm domega= bf v(P_0)cdotbf n .$$
But you can also set up an $epsilon/delta$-proof that does not make use of the MVT.
answered Jul 23 at 19:10
Christian Blatter
163k7107306
add a comment |Â
up vote
1
down vote
If the vector field $bf v$ is continuous in a neighborhood of $P_0$ then for any disc $D$ with center $P_0$ and unit normal $bf n$ one has
$$int_Dbf vcdotbf n>rm domega= bf v(P_*)cdotbf n rm area(D)$$
for some $P_*in D$. This follows from the standard MVT for continuous real-valued functions on a path-connected set $Dsubsetmathbb R^3$. It follows that
$$lim_r(D)to 0+1overrm area(D)int_Dbf vcdotbf n>rm domega= bf v(P_0)cdotbf n .$$
But you can also set up an $epsilon/delta$-proof that does not make use of the MVT.
answered Jul 23 at 19:10
Christian Blatter
163k7107306
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If the vector field $bf v$ is continuous in a neighborhood of $P_0$ then for any disc $D$ with center $P_0$ and unit normal $bf n$ one has
$$int_Dbf vcdotbf n>rm domega= bf v(P_*)cdotbf n rm area(D)$$
for some $P_*in D$. This follows from the standard MVT for continuous real-valued functions on a path-connected set $Dsubsetmathbb R^3$. It follows that
$$lim_r(D)to 0+1overrm area(D)int_Dbf vcdotbf n>rm domega= bf v(P_0)cdotbf n .$$
But you can also set up an $epsilon/delta$-proof that does not make use of the MVT.
answered Jul 23 at 19:10
Christian Blatter
163k7107306
If the vector field $bf v$ is continuous in a neighborhood of $P_0$ then for any disc $D$ with center $P_0$ and unit normal $bf n$ one has
$$int_Dbf vcdotbf n>rm domega= bf v(P_*)cdotbf n rm area(D)$$
for some $P_*in D$. This follows from the standard MVT for continuous real-valued functions on a path-connected set $Dsubsetmathbb R^3$. It follows that
$$lim_r(D)to 0+1overrm area(D)int_Dbf vcdotbf n>rm domega= bf v(P_0)cdotbf n .$$
But you can also set up an $epsilon/delta$-proof that does not make use of the MVT.
answered Jul 23 at 19:10
Christian Blatter
163k7107306
answered Jul 23 at 19:10
Christian Blatter
163k7107306
answered Jul 23 at 19:10
Christian Blatter
163k7107306
answered Jul 23 at 19:10
Christian Blatter
163k7107306
163k7107306
add a comment |Â
add a comment |Â
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The mean value theorem is in fact quite specific to one dimensional settings. In general, with several variables we get an inequality.
– Fimpellizieri
Jul 23 at 15:18
@Fimpellizieri I remember seeing a mean value theorem for double integrals in my textbook. Are you talking about the mean value theorem for derivatives?
– DWade64
Jul 23 at 15:21
We still don't get what you want with vector valued functions. The article linked has an example; see here.
– Fimpellizieri
Jul 23 at 15:27
@Fimpellizieri Ah that's interesting. Are there a class of functions which do satisfy the theorem? Because a surface integral is not much different than a double integral. It's still a volume over the surface where the height is given by $F_n$. It seems like it should work for some functions, but maybe not
– DWade64
Jul 23 at 15:50
Going from a single variable to multiple variables is a big change. I guess the most common but still powerful example is that things like saddles don't exist in one dimension.
– Fimpellizieri
Jul 23 at 15:57