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

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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










2












$begingroup$


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



enter image description here






vector-calculus







share|improve this question











$endgroup$











  • $begingroup$
    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♦
    2 hours 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
















2












$begingroup$


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



enter image description here






vector-calculus







share|improve this question











$endgroup$











  • $begingroup$
    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♦
    2 hours 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














2












2








2





$begingroup$


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



enter image description here






vector-calculus







share|improve this question











$endgroup$




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



enter image description here






vector-calculus




vector-calculus






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 1 hour ago









J. M. is slightly pensive♦

98.8k10311467




98.8k10311467










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♦
    2 hours 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






  • 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♦
    2 hours 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♦
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♦
2 hours 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











1 Answer
1






active

oldest

votes


















2












$begingroup$

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 *)





share|improve this answer









$endgroup$












  • $begingroup$
    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 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




    $begingroup$
    @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











Your Answer





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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

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 *)





share|improve this answer









$endgroup$












  • $begingroup$
    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 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




    $begingroup$
    @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















2












$begingroup$

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 *)





share|improve this answer









$endgroup$












  • $begingroup$
    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 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




    $begingroup$
    @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













2












2








2





$begingroup$

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 *)





share|improve this answer









$endgroup$



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 *)






share|improve this answer












share|improve this answer



share|improve this answer










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 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




    $begingroup$
    @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
















  • $begingroup$
    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 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




    $begingroup$
    @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















$begingroup$
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
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– Jose Enrique Calderon
1 hour ago




1




1




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@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





$begingroup$
@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] ]
$endgroup$
– Jose Enrique Calderon
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




1




$begingroup$
@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





$begingroup$
@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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