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vector calculus integration identity problem
The Next CEO of Stack Overflow$LaTeX$ format copy problemIs it possible to do vector calculus in Mathematica?Dipolar magnetic field lines inside a cylinderComparing unit normal definition in calculus with FrenetSerretSystemManipulating curl and div of a vector in spherical coordinatesIntegration with a matrix as the the integrandGet the vector Norm without absolute values?matrix calculus with types (similar to matrixcalculus.org)How do I verify a vector identity using Mathematica?Einstein summation convention for symbolic vector calculusVector calculus with index notation
$begingroup$
This is a follow up from another post . I was using the integration symbol available in the Basic Math Assistance available in Wolfram Mathematica.
I am new to vector calculus operations. There is a known identity found in the textbooks
$$int _4 pi hats (hatscdot A) d omega=frac4 pi3A$$
I have no idea how to do this type of integration. This is what I tried but return a dissaster
Integrate[s*(Dot[s, A]), s, 0, 4 [Pi]]
Also , without success
Integrate[Sin[[Theta]],
Cos[[Theta]]*(Dot[Sin[[Theta]], Cos[[Theta]], a1,
a2]), [Theta], 0, 4 [Pi]]
It is obviosu that I am doing something fundamentally not correct. I go to WM documentation on Vector Calculus but does not offer much substance or examples. How will you enter the equation above in order to return the identity in the right?
UPDATE 1
In respond to comment, here is a copy of the text. This is from page 10 Optical-Thermal Response of Laser-Irradiated Tissue ISBN 9789048188307
$$w$$ is the surface area of a sphere in solid angle steradian. s is the directional vector of a pencil of radiation located inside the sphere
vector-calculus
edited 1 hour ago
J. M. is slightly pensive♦
98.8k10311467
asked 2 hours ago
Jose Enrique CalderonJose Enrique Calderon
1,058718
$endgroup$
|
show 3 more comments
$begingroup$
This is a follow up from another post . I was using the integration symbol available in the Basic Math Assistance available in Wolfram Mathematica.
I am new to vector calculus operations. There is a known identity found in the textbooks
$$int _4 pi hats (hatscdot A) d omega=frac4 pi3A$$
I have no idea how to do this type of integration. This is what I tried but return a dissaster
Integrate[s*(Dot[s, A]), s, 0, 4 [Pi]]
Also , without success
Integrate[Sin[[Theta]],
Cos[[Theta]]*(Dot[Sin[[Theta]], Cos[[Theta]], a1,
a2]), [Theta], 0, 4 [Pi]]
It is obviosu that I am doing something fundamentally not correct. I go to WM documentation on Vector Calculus but does not offer much substance or examples. How will you enter the equation above in order to return the identity in the right?
UPDATE 1
In respond to comment, here is a copy of the text. This is from page 10 Optical-Thermal Response of Laser-Irradiated Tissue ISBN 9789048188307
$$w$$ is the surface area of a sphere in solid angle steradian. s is the directional vector of a pencil of radiation located inside the sphere
vector-calculus
edited 1 hour ago
J. M. is slightly pensive♦
98.8k10311467
asked 2 hours ago
Jose Enrique CalderonJose Enrique Calderon
1,058718
$endgroup$
$begingroup$
What are $s$ and $omega$ supposed to be? It might be helpful if you can give an example of the textbook with the formula.
$endgroup$
– J. M. is slightly pensive♦
2 hours ago
2
$begingroup$
Here's my guess:With[s = x, y, z, A = A1, A2, A3, Integrate[s (s.A), s [Element] Sphere[]] ]--- or this:With[s = x, y, z, A = A1, A2, A3, Integrate[s (s.A), s [Element] Sphere[]] == 4 Pi/3 A ]
$endgroup$
– Michael E2
2 hours ago
$begingroup$
@Michael, yes, that does seem to be it. This is why people should always define what their variables mean in their formulae.
$endgroup$
– J. M. is slightly pensive♦
1 hour ago
$begingroup$
@Michael E2 please post it as an answear for upvote
$endgroup$
– Jose Enrique Calderon
1 hour ago
1
$begingroup$
I've never seen this author's notation. My guess is that $int_4picdots$ means the integral over the sphere of measure $4pi$, i.e., the unit sphere.
$endgroup$
– Michael E2
1 hour ago
|
show 3 more comments
$begingroup$
This is a follow up from another post . I was using the integration symbol available in the Basic Math Assistance available in Wolfram Mathematica.
I am new to vector calculus operations. There is a known identity found in the textbooks
$$int _4 pi hats (hatscdot A) d omega=frac4 pi3A$$
I have no idea how to do this type of integration. This is what I tried but return a dissaster
Integrate[s*(Dot[s, A]), s, 0, 4 [Pi]]
Also , without success
Integrate[Sin[[Theta]],
Cos[[Theta]]*(Dot[Sin[[Theta]], Cos[[Theta]], a1,
a2]), [Theta], 0, 4 [Pi]]
It is obviosu that I am doing something fundamentally not correct. I go to WM documentation on Vector Calculus but does not offer much substance or examples. How will you enter the equation above in order to return the identity in the right?
UPDATE 1
In respond to comment, here is a copy of the text. This is from page 10 Optical-Thermal Response of Laser-Irradiated Tissue ISBN 9789048188307
$$w$$ is the surface area of a sphere in solid angle steradian. s is the directional vector of a pencil of radiation located inside the sphere
vector-calculus
edited 1 hour ago
J. M. is slightly pensive♦
98.8k10311467
asked 2 hours ago
Jose Enrique CalderonJose Enrique Calderon
1,058718
$endgroup$
This is a follow up from another post . I was using the integration symbol available in the Basic Math Assistance available in Wolfram Mathematica.
I am new to vector calculus operations. There is a known identity found in the textbooks
$$int _4 pi hats (hatscdot A) d omega=frac4 pi3A$$
I have no idea how to do this type of integration. This is what I tried but return a dissaster
Integrate[s*(Dot[s, A]), s, 0, 4 [Pi]]
Also , without success
Integrate[Sin[[Theta]],
Cos[[Theta]]*(Dot[Sin[[Theta]], Cos[[Theta]], a1,
a2]), [Theta], 0, 4 [Pi]]
It is obviosu that I am doing something fundamentally not correct. I go to WM documentation on Vector Calculus but does not offer much substance or examples. How will you enter the equation above in order to return the identity in the right?
UPDATE 1
In respond to comment, here is a copy of the text. This is from page 10 Optical-Thermal Response of Laser-Irradiated Tissue ISBN 9789048188307
$$w$$ is the surface area of a sphere in solid angle steradian. s is the directional vector of a pencil of radiation located inside the sphere
vector-calculus
vector-calculus
edited 1 hour ago
J. M. is slightly pensive♦
98.8k10311467
asked 2 hours ago
Jose Enrique CalderonJose Enrique Calderon
1,058718
edited 1 hour ago
J. M. is slightly pensive♦
98.8k10311467
asked 2 hours ago
Jose Enrique CalderonJose Enrique Calderon
1,058718
edited 1 hour ago
J. M. is slightly pensive♦
98.8k10311467
edited 1 hour ago
J. M. is slightly pensive♦
98.8k10311467
edited 1 hour ago
J. M. is slightly pensive♦
98.8k10311467
98.8k10311467
asked 2 hours ago
Jose Enrique CalderonJose Enrique Calderon
1,058718
asked 2 hours ago
Jose Enrique CalderonJose Enrique Calderon
1,058718
asked 2 hours ago
Jose Enrique CalderonJose Enrique Calderon
1,058718
1,058718
$begingroup$
What are $s$ and $omega$ supposed to be? It might be helpful if you can give an example of the textbook with the formula.
$endgroup$
– J. M. is slightly pensive♦
2 hours ago
2
$begingroup$
Here's my guess:With[s = x, y, z, A = A1, A2, A3, Integrate[s (s.A), s [Element] Sphere[]] ]--- or this:With[s = x, y, z, A = A1, A2, A3, Integrate[s (s.A), s [Element] Sphere[]] == 4 Pi/3 A ]
$endgroup$
– Michael E2
2 hours ago
$begingroup$
@Michael, yes, that does seem to be it. This is why people should always define what their variables mean in their formulae.
$endgroup$
– J. M. is slightly pensive♦
1 hour ago
$begingroup$
@Michael E2 please post it as an answear for upvote
$endgroup$
– Jose Enrique Calderon
1 hour ago
1
$begingroup$
I've never seen this author's notation. My guess is that $int_4picdots$ means the integral over the sphere of measure $4pi$, i.e., the unit sphere.
$endgroup$
– Michael E2
1 hour ago
|
show 3 more comments
$begingroup$
What are $s$ and $omega$ supposed to be? It might be helpful if you can give an example of the textbook with the formula.
$endgroup$
– J. M. is slightly pensive♦
2 hours ago
2
$begingroup$
Here's my guess:With[s = x, y, z, A = A1, A2, A3, Integrate[s (s.A), s [Element] Sphere[]] ]--- or this:With[s = x, y, z, A = A1, A2, A3, Integrate[s (s.A), s [Element] Sphere[]] == 4 Pi/3 A ]
$endgroup$
– Michael E2
2 hours ago
$begingroup$
@Michael, yes, that does seem to be it. This is why people should always define what their variables mean in their formulae.
$endgroup$
– J. M. is slightly pensive♦
1 hour ago
$begingroup$
@Michael E2 please post it as an answear for upvote
$endgroup$
– Jose Enrique Calderon
1 hour ago
1
$begingroup$
I've never seen this author's notation. My guess is that $int_4picdots$ means the integral over the sphere of measure $4pi$, i.e., the unit sphere.
$endgroup$
– Michael E2
1 hour ago
$begingroup$
What are $s$ and $omega$ supposed to be? It might be helpful if you can give an example of the textbook with the formula.
$endgroup$
– J. M. is slightly pensive♦
2 hours ago
$begingroup$
What are $s$ and $omega$ supposed to be? It might be helpful if you can give an example of the textbook with the formula.
$endgroup$
– J. M. is slightly pensive♦
2 hours ago
2
2
$begingroup$
Here's my guess:
With[s = x, y, z, A = A1, A2, A3, Integrate[s (s.A), s [Element] Sphere[]] ] --- or this: With[s = x, y, z, A = A1, A2, A3, Integrate[s (s.A), s [Element] Sphere[]] == 4 Pi/3 A ]$endgroup$
– Michael E2
2 hours ago
$begingroup$
Here's my guess:
With[s = x, y, z, A = A1, A2, A3, Integrate[s (s.A), s [Element] Sphere[]] ] --- or this: With[s = x, y, z, A = A1, A2, A3, Integrate[s (s.A), s [Element] Sphere[]] == 4 Pi/3 A ]$endgroup$
– Michael E2
2 hours ago
$begingroup$
@Michael, yes, that does seem to be it. This is why people should always define what their variables mean in their formulae.
$endgroup$
– J. M. is slightly pensive♦
1 hour ago
$begingroup$
@Michael, yes, that does seem to be it. This is why people should always define what their variables mean in their formulae.
$endgroup$
– J. M. is slightly pensive♦
1 hour ago
$begingroup$
@Michael E2 please post it as an answear for upvote
$endgroup$
– Jose Enrique Calderon
1 hour ago
$begingroup$
@Michael E2 please post it as an answear for upvote
$endgroup$
– Jose Enrique Calderon
1 hour ago
1
1
$begingroup$
I've never seen this author's notation. My guess is that $int_4picdots$ means the integral over the sphere of measure $4pi$, i.e., the unit sphere.
$endgroup$
– Michael E2
1 hour ago
$begingroup$
I've never seen this author's notation. My guess is that $int_4picdots$ means the integral over the sphere of measure $4pi$, i.e., the unit sphere.
$endgroup$
– Michael E2
1 hour ago
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Here's my guess:
With[s = x, y, z,
A = A1, A2, A3, Integrate[s (s.A), s ∈ Sphere[]] ]
(* (4 A1 π)/3, (4 A2 π)/3, (4 A3 π)/3 *)
--- or this:
With[s = x, y, z, A = A1, A2, A3,
Integrate[s (s.A), s ∈ Sphere[]] == 4 Pi/3 A ]
(* True *)
answered 1 hour ago
Michael E2Michael E2
150k12203482
$endgroup$
$begingroup$
Why it simply does not work with limits of integration s,0,4Pi
$endgroup$
– Jose Enrique Calderon
1 hour ago
1
$begingroup$
@Jose, the author was being lazy with the limits (basically, shorter than saying "integrate over the whole area of the unit sphere"). It is fine to be lazy in mathematics, but not so much when programming.
$endgroup$
– J. M. is slightly pensive♦
1 hour ago
$begingroup$
@J.M. is slightly pensive Ok.. but why Mathematica function proposed in the answear does not work with With[s = x, y, z, A = A1, A2, A3, Integrate[s (s.A), s,0,4Pi] ]
$endgroup$
– Jose Enrique Calderon
1 hour ago
1
$begingroup$
@Jose The syntaxs, 0, 4 Pialready implies one-dimensionalsfrom Mathematica's view, while in the "abuse of notation" used in your reference, $hats$ is implied to be a vector.
$endgroup$
– J. M. is slightly pensive♦
1 hour ago
1
$begingroup$
@Jose: the easiest way is that you have to switch to spherical coordinates if you need to integrate across arbitrary angles. If you insist on keeping yourself to regions, you can useRegionIntersection[]withSphere[]and eitherConicHullRegion[]orHalfSpace[].
$endgroup$
– J. M. is slightly pensive♦
1 hour ago
|
show 2 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here's my guess:
With[s = x, y, z,
A = A1, A2, A3, Integrate[s (s.A), s ∈ Sphere[]] ]
(* (4 A1 π)/3, (4 A2 π)/3, (4 A3 π)/3 *)
--- or this:
With[s = x, y, z, A = A1, A2, A3,
Integrate[s (s.A), s ∈ Sphere[]] == 4 Pi/3 A ]
(* True *)
answered 1 hour ago
Michael E2Michael E2
150k12203482
$endgroup$
$begingroup$
Why it simply does not work with limits of integration s,0,4Pi
$endgroup$
– Jose Enrique Calderon
1 hour ago
1
$begingroup$
@Jose, the author was being lazy with the limits (basically, shorter than saying "integrate over the whole area of the unit sphere"). It is fine to be lazy in mathematics, but not so much when programming.
$endgroup$
– J. M. is slightly pensive♦
1 hour ago
$begingroup$
@J.M. is slightly pensive Ok.. but why Mathematica function proposed in the answear does not work with With[s = x, y, z, A = A1, A2, A3, Integrate[s (s.A), s,0,4Pi] ]
$endgroup$
– Jose Enrique Calderon
1 hour ago
1
$begingroup$
@Jose The syntaxs, 0, 4 Pialready implies one-dimensionalsfrom Mathematica's view, while in the "abuse of notation" used in your reference, $hats$ is implied to be a vector.
$endgroup$
– J. M. is slightly pensive♦
1 hour ago
1
$begingroup$
@Jose: the easiest way is that you have to switch to spherical coordinates if you need to integrate across arbitrary angles. If you insist on keeping yourself to regions, you can useRegionIntersection[]withSphere[]and eitherConicHullRegion[]orHalfSpace[].
$endgroup$
– J. M. is slightly pensive♦
1 hour ago
|
show 2 more comments
$begingroup$
Here's my guess:
With[s = x, y, z,
A = A1, A2, A3, Integrate[s (s.A), s ∈ Sphere[]] ]
(* (4 A1 π)/3, (4 A2 π)/3, (4 A3 π)/3 *)
--- or this:
With[s = x, y, z, A = A1, A2, A3,
Integrate[s (s.A), s ∈ Sphere[]] == 4 Pi/3 A ]
(* True *)
answered 1 hour ago
Michael E2Michael E2
150k12203482
$endgroup$
$begingroup$
Why it simply does not work with limits of integration s,0,4Pi
$endgroup$
– Jose Enrique Calderon
1 hour ago
1
$begingroup$
@Jose, the author was being lazy with the limits (basically, shorter than saying "integrate over the whole area of the unit sphere"). It is fine to be lazy in mathematics, but not so much when programming.
$endgroup$
– J. M. is slightly pensive♦
1 hour ago
$begingroup$
@J.M. is slightly pensive Ok.. but why Mathematica function proposed in the answear does not work with With[s = x, y, z, A = A1, A2, A3, Integrate[s (s.A), s,0,4Pi] ]
$endgroup$
– Jose Enrique Calderon
1 hour ago
1
$begingroup$
@Jose The syntaxs, 0, 4 Pialready implies one-dimensionalsfrom Mathematica's view, while in the "abuse of notation" used in your reference, $hats$ is implied to be a vector.
$endgroup$
– J. M. is slightly pensive♦
1 hour ago
1
$begingroup$
@Jose: the easiest way is that you have to switch to spherical coordinates if you need to integrate across arbitrary angles. If you insist on keeping yourself to regions, you can useRegionIntersection[]withSphere[]and eitherConicHullRegion[]orHalfSpace[].
$endgroup$
– J. M. is slightly pensive♦
1 hour ago
|
show 2 more comments
$begingroup$
Here's my guess:
With[s = x, y, z,
A = A1, A2, A3, Integrate[s (s.A), s ∈ Sphere[]] ]
(* (4 A1 π)/3, (4 A2 π)/3, (4 A3 π)/3 *)
--- or this:
With[s = x, y, z, A = A1, A2, A3,
Integrate[s (s.A), s ∈ Sphere[]] == 4 Pi/3 A ]
(* True *)
answered 1 hour ago
Michael E2Michael E2
150k12203482
$endgroup$
Here's my guess:
With[s = x, y, z,
A = A1, A2, A3, Integrate[s (s.A), s ∈ Sphere[]] ]
(* (4 A1 π)/3, (4 A2 π)/3, (4 A3 π)/3 *)
--- or this:
With[s = x, y, z, A = A1, A2, A3,
Integrate[s (s.A), s ∈ Sphere[]] == 4 Pi/3 A ]
(* True *)
answered 1 hour ago
Michael E2Michael E2
150k12203482
answered 1 hour ago
Michael E2Michael E2
150k12203482
answered 1 hour ago
Michael E2Michael E2
150k12203482
answered 1 hour ago
Michael E2Michael E2
150k12203482
150k12203482
$begingroup$
Why it simply does not work with limits of integration s,0,4Pi
$endgroup$
– Jose Enrique Calderon
1 hour ago
1
$begingroup$
@Jose, the author was being lazy with the limits (basically, shorter than saying "integrate over the whole area of the unit sphere"). It is fine to be lazy in mathematics, but not so much when programming.
$endgroup$
– J. M. is slightly pensive♦
1 hour ago
$begingroup$
@J.M. is slightly pensive Ok.. but why Mathematica function proposed in the answear does not work with With[s = x, y, z, A = A1, A2, A3, Integrate[s (s.A), s,0,4Pi] ]
$endgroup$
– Jose Enrique Calderon
1 hour ago
1
$begingroup$
@Jose The syntaxs, 0, 4 Pialready implies one-dimensionalsfrom Mathematica's view, while in the "abuse of notation" used in your reference, $hats$ is implied to be a vector.
$endgroup$
– J. M. is slightly pensive♦
1 hour ago
1
$begingroup$
@Jose: the easiest way is that you have to switch to spherical coordinates if you need to integrate across arbitrary angles. If you insist on keeping yourself to regions, you can useRegionIntersection[]withSphere[]and eitherConicHullRegion[]orHalfSpace[].
$endgroup$
– J. M. is slightly pensive♦
1 hour ago
|
show 2 more comments
$begingroup$
Why it simply does not work with limits of integration s,0,4Pi
$endgroup$
– Jose Enrique Calderon
1 hour ago
1
$begingroup$
@Jose, the author was being lazy with the limits (basically, shorter than saying "integrate over the whole area of the unit sphere"). It is fine to be lazy in mathematics, but not so much when programming.
$endgroup$
– J. M. is slightly pensive♦
1 hour ago
$begingroup$
@J.M. is slightly pensive Ok.. but why Mathematica function proposed in the answear does not work with With[s = x, y, z, A = A1, A2, A3, Integrate[s (s.A), s,0,4Pi] ]
$endgroup$
– Jose Enrique Calderon
1 hour ago
1
$begingroup$
@Jose The syntaxs, 0, 4 Pialready implies one-dimensionalsfrom Mathematica's view, while in the "abuse of notation" used in your reference, $hats$ is implied to be a vector.
$endgroup$
– J. M. is slightly pensive♦
1 hour ago
1
$begingroup$
@Jose: the easiest way is that you have to switch to spherical coordinates if you need to integrate across arbitrary angles. If you insist on keeping yourself to regions, you can useRegionIntersection[]withSphere[]and eitherConicHullRegion[]orHalfSpace[].
$endgroup$
– J. M. is slightly pensive♦
1 hour ago
$begingroup$
Why it simply does not work with limits of integration s,0,4Pi
$endgroup$
– Jose Enrique Calderon
1 hour ago
$begingroup$
Why it simply does not work with limits of integration s,0,4Pi
$endgroup$
– Jose Enrique Calderon
1 hour ago
1
1
$begingroup$
@Jose, the author was being lazy with the limits (basically, shorter than saying "integrate over the whole area of the unit sphere"). It is fine to be lazy in mathematics, but not so much when programming.
$endgroup$
– J. M. is slightly pensive♦
1 hour ago
$begingroup$
@Jose, the author was being lazy with the limits (basically, shorter than saying "integrate over the whole area of the unit sphere"). It is fine to be lazy in mathematics, but not so much when programming.
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– J. M. is slightly pensive♦
1 hour ago
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@J.M. is slightly pensive Ok.. but why Mathematica function proposed in the answear does not work with With[s = x, y, z, A = A1, A2, A3, Integrate[s (s.A), s,0,4Pi] ]
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– Jose Enrique Calderon
1 hour ago
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@J.M. is slightly pensive Ok.. but why Mathematica function proposed in the answear does not work with With[s = x, y, z, A = A1, A2, A3, Integrate[s (s.A), s,0,4Pi] ]
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– Jose Enrique Calderon
1 hour ago
1
1
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@Jose The syntax
s, 0, 4 Pi already implies one-dimensional s from Mathematica's view, while in the "abuse of notation" used in your reference, $hats$ is implied to be a vector.$endgroup$
– J. M. is slightly pensive♦
1 hour ago
$begingroup$
@Jose The syntax
s, 0, 4 Pi already implies one-dimensional s from Mathematica's view, while in the "abuse of notation" used in your reference, $hats$ is implied to be a vector.$endgroup$
– J. M. is slightly pensive♦
1 hour ago
1
1
$begingroup$
@Jose: the easiest way is that you have to switch to spherical coordinates if you need to integrate across arbitrary angles. If you insist on keeping yourself to regions, you can use
RegionIntersection[] with Sphere[] and either ConicHullRegion[] or HalfSpace[].$endgroup$
– J. M. is slightly pensive♦
1 hour ago
$begingroup$
@Jose: the easiest way is that you have to switch to spherical coordinates if you need to integrate across arbitrary angles. If you insist on keeping yourself to regions, you can use
RegionIntersection[] with Sphere[] and either ConicHullRegion[] or HalfSpace[].$endgroup$
– J. M. is slightly pensive♦
1 hour ago
|
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$begingroup$
What are $s$ and $omega$ supposed to be? It might be helpful if you can give an example of the textbook with the formula.
$endgroup$
– J. M. is slightly pensive♦
2 hours ago
2
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Here's my guess:
With[s = x, y, z, A = A1, A2, A3, Integrate[s (s.A), s [Element] Sphere[]] ]--- or this:With[s = x, y, z, A = A1, A2, A3, Integrate[s (s.A), s [Element] Sphere[]] == 4 Pi/3 A ]$endgroup$
– Michael E2
2 hours ago
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@Michael, yes, that does seem to be it. This is why people should always define what their variables mean in their formulae.
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– J. M. is slightly pensive♦
1 hour ago
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@Michael E2 please post it as an answear for upvote
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– Jose Enrique Calderon
1 hour ago
1
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I've never seen this author's notation. My guess is that $int_4picdots$ means the integral over the sphere of measure $4pi$, i.e., the unit sphere.
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– Michael E2
1 hour ago