Mixing
Integration over a spherical surface
0
$begingroup$
Suppose that we have let's say a function of "something" that is everywhere on the spherical surface zero except at one point it is finite. Why is the integral of such a function over the surface is zero?
integration
asked Dec 8 '14 at 23:21
user198714user198714
31
$endgroup$
add a comment |
0
$begingroup$
Suppose that we have let's say a function of "something" that is everywhere on the spherical surface zero except at one point it is finite. Why is the integral of such a function over the surface is zero?
integration
asked Dec 8 '14 at 23:21
user198714user198714
31
$endgroup$
$begingroup$
You mean you have a (discontinuous) function $f:S^2to mathbb{R}$ such that $f(z)=0$ except for one value of $z$?
$endgroup$
– rogerl
Dec 8 '14 at 23:27
$begingroup$
I don't know about the S^2 --> R notation but the function is everywhere zero on the shpere except at one point. @rogerl
$endgroup$
– user198714
Dec 9 '14 at 9:10
add a comment |
0
0
0
$begingroup$
Suppose that we have let's say a function of "something" that is everywhere on the spherical surface zero except at one point it is finite. Why is the integral of such a function over the surface is zero?
integration
asked Dec 8 '14 at 23:21
user198714user198714
31
$endgroup$
Suppose that we have let's say a function of "something" that is everywhere on the spherical surface zero except at one point it is finite. Why is the integral of such a function over the surface is zero?
integration
integration
asked Dec 8 '14 at 23:21
user198714user198714
31
asked Dec 8 '14 at 23:21
user198714user198714
31
asked Dec 8 '14 at 23:21
user198714user198714
31
asked Dec 8 '14 at 23:21
user198714user198714
31
asked Dec 8 '14 at 23:21
user198714user198714
31
31
$begingroup$
You mean you have a (discontinuous) function $f:S^2to mathbb{R}$ such that $f(z)=0$ except for one value of $z$?
$endgroup$
– rogerl
Dec 8 '14 at 23:27
$begingroup$
I don't know about the S^2 --> R notation but the function is everywhere zero on the shpere except at one point. @rogerl
$endgroup$
– user198714
Dec 9 '14 at 9:10
add a comment |
$begingroup$
You mean you have a (discontinuous) function $f:S^2to mathbb{R}$ such that $f(z)=0$ except for one value of $z$?
$endgroup$
– rogerl
Dec 8 '14 at 23:27
$begingroup$
I don't know about the S^2 --> R notation but the function is everywhere zero on the shpere except at one point. @rogerl
$endgroup$
– user198714
Dec 9 '14 at 9:10
$begingroup$
You mean you have a (discontinuous) function $f:S^2to mathbb{R}$ such that $f(z)=0$ except for one value of $z$?
$endgroup$
– rogerl
Dec 8 '14 at 23:27
$begingroup$
You mean you have a (discontinuous) function $f:S^2to mathbb{R}$ such that $f(z)=0$ except for one value of $z$?
$endgroup$
– rogerl
Dec 8 '14 at 23:27
$begingroup$
I don't know about the S^2 --> R notation but the function is everywhere zero on the shpere except at one point. @rogerl
$endgroup$
– user198714
Dec 9 '14 at 9:10
$begingroup$
I don't know about the S^2 --> R notation but the function is everywhere zero on the shpere except at one point. @rogerl
$endgroup$
– user198714
Dec 9 '14 at 9:10
add a comment |
2 Answers
2
active
oldest
votes
1
$begingroup$
The set where the integrand is not zero is a zero set. Therefore the integral vanishes.
answered Dec 8 '14 at 23:27
Loreno HeerLoreno Heer
3,34411534
$endgroup$
$begingroup$
May you please elaborate? I don't know a lot about sets. I probably should have posted this is the Physics section?
$endgroup$
– user198714
Dec 9 '14 at 9:11
$begingroup$
Just note that in the definition of the Riemann-sum, changing the integrand (the function you try to integrate) on just one point does not change anything (in the limit): $sum_{i=1}^n f(t_i) (x_i - x_{i-1})$
$endgroup$
– Loreno Heer
Dec 9 '14 at 15:12
$begingroup$
Aha but here we are talking not on just one point, we are intergating over the whole sphere. Sorry if I am not following you quickly. Does this have a proof or something I can look at? @sanjab
$endgroup$
– user198714
Dec 9 '14 at 15:54
add a comment |
0
$begingroup$
Suppose $f(x)$ is identically zero in $[a,b]$ expect at some point $c$ in between then the definite integral from $a$ to $b$ would vanish. Your question is analogous to this example. The example which you said is taking place in higher dimensions so we are not able to visualise it. Hope you get my point.
edited Jan 18 at 7:08
dantopa
6,53942244
answered Jan 17 at 22:15
user566574user566574
154
$endgroup$
$begingroup$
Welcome to MSE. Your answer adds nothing new to the already existing (and accepted) answer.
$endgroup$
– José Carlos Santos
Jan 17 at 22:47
$begingroup$
@JoséCarlosSantos I feel like I should point out that the OP was confused by the phrasing of the accepted answer. It's possible that one dimensional integrals are more familiar to the OP, and thus that this answer will be more helpful in providing insight.
$endgroup$
– jgon
Jan 18 at 0:52
$begingroup$
Welcome to Mathematics Stack Exchange! Take the short tour to see how how to get the most from your time here. For typesetting equations we use MathJax.
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– dantopa
Jan 18 at 7:07
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
$begingroup$
The set where the integrand is not zero is a zero set. Therefore the integral vanishes.
answered Dec 8 '14 at 23:27
Loreno HeerLoreno Heer
3,34411534
$endgroup$
$begingroup$
May you please elaborate? I don't know a lot about sets. I probably should have posted this is the Physics section?
$endgroup$
– user198714
Dec 9 '14 at 9:11
$begingroup$
Just note that in the definition of the Riemann-sum, changing the integrand (the function you try to integrate) on just one point does not change anything (in the limit): $sum_{i=1}^n f(t_i) (x_i - x_{i-1})$
$endgroup$
– Loreno Heer
Dec 9 '14 at 15:12
$begingroup$
Aha but here we are talking not on just one point, we are intergating over the whole sphere. Sorry if I am not following you quickly. Does this have a proof or something I can look at? @sanjab
$endgroup$
– user198714
Dec 9 '14 at 15:54
add a comment |
1
$begingroup$
The set where the integrand is not zero is a zero set. Therefore the integral vanishes.
answered Dec 8 '14 at 23:27
Loreno HeerLoreno Heer
3,34411534
$endgroup$
$begingroup$
May you please elaborate? I don't know a lot about sets. I probably should have posted this is the Physics section?
$endgroup$
– user198714
Dec 9 '14 at 9:11
$begingroup$
Just note that in the definition of the Riemann-sum, changing the integrand (the function you try to integrate) on just one point does not change anything (in the limit): $sum_{i=1}^n f(t_i) (x_i - x_{i-1})$
$endgroup$
– Loreno Heer
Dec 9 '14 at 15:12
$begingroup$
Aha but here we are talking not on just one point, we are intergating over the whole sphere. Sorry if I am not following you quickly. Does this have a proof or something I can look at? @sanjab
$endgroup$
– user198714
Dec 9 '14 at 15:54
add a comment |
1
1
1
$begingroup$
The set where the integrand is not zero is a zero set. Therefore the integral vanishes.
answered Dec 8 '14 at 23:27
Loreno HeerLoreno Heer
3,34411534
$endgroup$
The set where the integrand is not zero is a zero set. Therefore the integral vanishes.
answered Dec 8 '14 at 23:27
Loreno HeerLoreno Heer
3,34411534
answered Dec 8 '14 at 23:27
Loreno HeerLoreno Heer
3,34411534
answered Dec 8 '14 at 23:27
Loreno HeerLoreno Heer
3,34411534
answered Dec 8 '14 at 23:27
Loreno HeerLoreno Heer
3,34411534
3,34411534
$begingroup$
May you please elaborate? I don't know a lot about sets. I probably should have posted this is the Physics section?
$endgroup$
– user198714
Dec 9 '14 at 9:11
$begingroup$
Just note that in the definition of the Riemann-sum, changing the integrand (the function you try to integrate) on just one point does not change anything (in the limit): $sum_{i=1}^n f(t_i) (x_i - x_{i-1})$
$endgroup$
– Loreno Heer
Dec 9 '14 at 15:12
$begingroup$
Aha but here we are talking not on just one point, we are intergating over the whole sphere. Sorry if I am not following you quickly. Does this have a proof or something I can look at? @sanjab
$endgroup$
– user198714
Dec 9 '14 at 15:54
add a comment |
$begingroup$
May you please elaborate? I don't know a lot about sets. I probably should have posted this is the Physics section?
$endgroup$
– user198714
Dec 9 '14 at 9:11
$begingroup$
Just note that in the definition of the Riemann-sum, changing the integrand (the function you try to integrate) on just one point does not change anything (in the limit): $sum_{i=1}^n f(t_i) (x_i - x_{i-1})$
$endgroup$
– Loreno Heer
Dec 9 '14 at 15:12
$begingroup$
Aha but here we are talking not on just one point, we are intergating over the whole sphere. Sorry if I am not following you quickly. Does this have a proof or something I can look at? @sanjab
$endgroup$
– user198714
Dec 9 '14 at 15:54
$begingroup$
May you please elaborate? I don't know a lot about sets. I probably should have posted this is the Physics section?
$endgroup$
– user198714
Dec 9 '14 at 9:11
$begingroup$
May you please elaborate? I don't know a lot about sets. I probably should have posted this is the Physics section?
$endgroup$
– user198714
Dec 9 '14 at 9:11
$begingroup$
Just note that in the definition of the Riemann-sum, changing the integrand (the function you try to integrate) on just one point does not change anything (in the limit): $sum_{i=1}^n f(t_i) (x_i - x_{i-1})$
$endgroup$
– Loreno Heer
Dec 9 '14 at 15:12
$begingroup$
Just note that in the definition of the Riemann-sum, changing the integrand (the function you try to integrate) on just one point does not change anything (in the limit): $sum_{i=1}^n f(t_i) (x_i - x_{i-1})$
$endgroup$
– Loreno Heer
Dec 9 '14 at 15:12
$begingroup$
Aha but here we are talking not on just one point, we are intergating over the whole sphere. Sorry if I am not following you quickly. Does this have a proof or something I can look at? @sanjab
$endgroup$
– user198714
Dec 9 '14 at 15:54
$begingroup$
Aha but here we are talking not on just one point, we are intergating over the whole sphere. Sorry if I am not following you quickly. Does this have a proof or something I can look at? @sanjab
$endgroup$
– user198714
Dec 9 '14 at 15:54
add a comment |
0
$begingroup$
Suppose $f(x)$ is identically zero in $[a,b]$ expect at some point $c$ in between then the definite integral from $a$ to $b$ would vanish. Your question is analogous to this example. The example which you said is taking place in higher dimensions so we are not able to visualise it. Hope you get my point.
edited Jan 18 at 7:08
dantopa
6,53942244
answered Jan 17 at 22:15
user566574user566574
154
$endgroup$
$begingroup$
Welcome to MSE. Your answer adds nothing new to the already existing (and accepted) answer.
$endgroup$
– José Carlos Santos
Jan 17 at 22:47
$begingroup$
@JoséCarlosSantos I feel like I should point out that the OP was confused by the phrasing of the accepted answer. It's possible that one dimensional integrals are more familiar to the OP, and thus that this answer will be more helpful in providing insight.
$endgroup$
– jgon
Jan 18 at 0:52
$begingroup$
Welcome to Mathematics Stack Exchange! Take the short tour to see how how to get the most from your time here. For typesetting equations we use MathJax.
$endgroup$
– dantopa
Jan 18 at 7:07
add a comment |
0
$begingroup$
Suppose $f(x)$ is identically zero in $[a,b]$ expect at some point $c$ in between then the definite integral from $a$ to $b$ would vanish. Your question is analogous to this example. The example which you said is taking place in higher dimensions so we are not able to visualise it. Hope you get my point.
edited Jan 18 at 7:08
dantopa
6,53942244
answered Jan 17 at 22:15
user566574user566574
154
$endgroup$
$begingroup$
Welcome to MSE. Your answer adds nothing new to the already existing (and accepted) answer.
$endgroup$
– José Carlos Santos
Jan 17 at 22:47
$begingroup$
@JoséCarlosSantos I feel like I should point out that the OP was confused by the phrasing of the accepted answer. It's possible that one dimensional integrals are more familiar to the OP, and thus that this answer will be more helpful in providing insight.
$endgroup$
– jgon
Jan 18 at 0:52
$begingroup$
Welcome to Mathematics Stack Exchange! Take the short tour to see how how to get the most from your time here. For typesetting equations we use MathJax.
$endgroup$
– dantopa
Jan 18 at 7:07
add a comment |
0
0
0
$begingroup$
Suppose $f(x)$ is identically zero in $[a,b]$ expect at some point $c$ in between then the definite integral from $a$ to $b$ would vanish. Your question is analogous to this example. The example which you said is taking place in higher dimensions so we are not able to visualise it. Hope you get my point.
edited Jan 18 at 7:08
dantopa
6,53942244
answered Jan 17 at 22:15
user566574user566574
154
$endgroup$
Suppose $f(x)$ is identically zero in $[a,b]$ expect at some point $c$ in between then the definite integral from $a$ to $b$ would vanish. Your question is analogous to this example. The example which you said is taking place in higher dimensions so we are not able to visualise it. Hope you get my point.
edited Jan 18 at 7:08
dantopa
6,53942244
answered Jan 17 at 22:15
user566574user566574
154
edited Jan 18 at 7:08
dantopa
6,53942244
edited Jan 18 at 7:08
dantopa
6,53942244
edited Jan 18 at 7:08
dantopa
6,53942244
6,53942244
answered Jan 17 at 22:15
user566574user566574
154
answered Jan 17 at 22:15
user566574user566574
154
answered Jan 17 at 22:15
user566574user566574
154
154
$begingroup$
Welcome to MSE. Your answer adds nothing new to the already existing (and accepted) answer.
$endgroup$
– José Carlos Santos
Jan 17 at 22:47
$begingroup$
@JoséCarlosSantos I feel like I should point out that the OP was confused by the phrasing of the accepted answer. It's possible that one dimensional integrals are more familiar to the OP, and thus that this answer will be more helpful in providing insight.
$endgroup$
– jgon
Jan 18 at 0:52
$begingroup$
Welcome to Mathematics Stack Exchange! Take the short tour to see how how to get the most from your time here. For typesetting equations we use MathJax.
$endgroup$
– dantopa
Jan 18 at 7:07
add a comment |
$begingroup$
Welcome to MSE. Your answer adds nothing new to the already existing (and accepted) answer.
$endgroup$
– José Carlos Santos
Jan 17 at 22:47
$begingroup$
@JoséCarlosSantos I feel like I should point out that the OP was confused by the phrasing of the accepted answer. It's possible that one dimensional integrals are more familiar to the OP, and thus that this answer will be more helpful in providing insight.
$endgroup$
– jgon
Jan 18 at 0:52
$begingroup$
Welcome to Mathematics Stack Exchange! Take the short tour to see how how to get the most from your time here. For typesetting equations we use MathJax.
$endgroup$
– dantopa
Jan 18 at 7:07
$begingroup$
Welcome to MSE. Your answer adds nothing new to the already existing (and accepted) answer.
$endgroup$
– José Carlos Santos
Jan 17 at 22:47
$begingroup$
Welcome to MSE. Your answer adds nothing new to the already existing (and accepted) answer.
$endgroup$
– José Carlos Santos
Jan 17 at 22:47
$begingroup$
@JoséCarlosSantos I feel like I should point out that the OP was confused by the phrasing of the accepted answer. It's possible that one dimensional integrals are more familiar to the OP, and thus that this answer will be more helpful in providing insight.
$endgroup$
– jgon
Jan 18 at 0:52
$begingroup$
@JoséCarlosSantos I feel like I should point out that the OP was confused by the phrasing of the accepted answer. It's possible that one dimensional integrals are more familiar to the OP, and thus that this answer will be more helpful in providing insight.
$endgroup$
– jgon
Jan 18 at 0:52
$begingroup$
Welcome to Mathematics Stack Exchange! Take the short tour to see how how to get the most from your time here. For typesetting equations we use MathJax.
$endgroup$
– dantopa
Jan 18 at 7:07
$begingroup$
Welcome to Mathematics Stack Exchange! Take the short tour to see how how to get the most from your time here. For typesetting equations we use MathJax.
$endgroup$
– dantopa
Jan 18 at 7:07
add a comment |
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$begingroup$
You mean you have a (discontinuous) function $f:S^2to mathbb{R}$ such that $f(z)=0$ except for one value of $z$?
$endgroup$
– rogerl
Dec 8 '14 at 23:27
$begingroup$
I don't know about the S^2 --> R notation but the function is everywhere zero on the shpere except at one point. @rogerl
$endgroup$
– user198714
Dec 9 '14 at 9:10