Vector calculus integration identity problem












3












$begingroup$


This is a follow up from another post. I was using the integration symbols available in the Basic Math Assistant palette.



I am new to vector calculus operations. There is a known identity found in my textbook.



$$qquad int _{4 pi }hat{s} (hat{s}cdot A) d omega=frac{4 pi}{3}A$$



I have no idea how to do this type of integration. This is what I tried, but it returns a disaster:



Integrate[s*(Dot[s, A]), s, {0, 4 π}]


Also without success:



Integrate[{Sin[θ], Cos[θ]}*(Dot[{Sin[θ], Cos[θ]}, {a1, a2}]), θ, {0, 4 π}]


It is obvious that I am doing something fundamentally not correct. I go to the documentation on Vector Calculus, but it does not offer much in substance or examples. How do you enter the integral expression shown above in order to return the identity in the right?



Update



In response to comments, here is a copy of the text. This is from page 10 of Optical-Thermal Response of Laser-Irradiated Tissue.



$omega$ is the surface area of a sphere in steradians. $hat s$ is the directional vector of a pencil of radiation located inside the sphere



table






symbolic 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♦
    3 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
    2 hours ago








  • 1




    $begingroup$
    I've never seen this author's notation. My guess is that $int_{4pi}cdots$ means the integral over the sphere of measure $4pi$, i.e., the unit sphere.
    $endgroup$
    – Michael E2
    2 hours ago


















3












$begingroup$


This is a follow up from another post. I was using the integration symbols available in the Basic Math Assistant palette.



I am new to vector calculus operations. There is a known identity found in my textbook.



$$qquad int _{4 pi }hat{s} (hat{s}cdot A) d omega=frac{4 pi}{3}A$$



I have no idea how to do this type of integration. This is what I tried, but it returns a disaster:



Integrate[s*(Dot[s, A]), s, {0, 4 π}]


Also without success:



Integrate[{Sin[θ], Cos[θ]}*(Dot[{Sin[θ], Cos[θ]}, {a1, a2}]), θ, {0, 4 π}]


It is obvious that I am doing something fundamentally not correct. I go to the documentation on Vector Calculus, but it does not offer much in substance or examples. How do you enter the integral expression shown above in order to return the identity in the right?



Update



In response to comments, here is a copy of the text. This is from page 10 of Optical-Thermal Response of Laser-Irradiated Tissue.



$omega$ is the surface area of a sphere in steradians. $hat s$ is the directional vector of a pencil of radiation located inside the sphere



table






symbolic 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♦
    3 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
    2 hours ago








  • 1




    $begingroup$
    I've never seen this author's notation. My guess is that $int_{4pi}cdots$ means the integral over the sphere of measure $4pi$, i.e., the unit sphere.
    $endgroup$
    – Michael E2
    2 hours ago
















3












3








3





$begingroup$


This is a follow up from another post. I was using the integration symbols available in the Basic Math Assistant palette.



I am new to vector calculus operations. There is a known identity found in my textbook.



$$qquad int _{4 pi }hat{s} (hat{s}cdot A) d omega=frac{4 pi}{3}A$$



I have no idea how to do this type of integration. This is what I tried, but it returns a disaster:



Integrate[s*(Dot[s, A]), s, {0, 4 π}]


Also without success:



Integrate[{Sin[θ], Cos[θ]}*(Dot[{Sin[θ], Cos[θ]}, {a1, a2}]), θ, {0, 4 π}]


It is obvious that I am doing something fundamentally not correct. I go to the documentation on Vector Calculus, but it does not offer much in substance or examples. How do you enter the integral expression shown above in order to return the identity in the right?



Update



In response to comments, here is a copy of the text. This is from page 10 of Optical-Thermal Response of Laser-Irradiated Tissue.



$omega$ is the surface area of a sphere in steradians. $hat s$ is the directional vector of a pencil of radiation located inside the sphere



table






symbolic vector-calculus







share|improve this question











$endgroup$




This is a follow up from another post. I was using the integration symbols available in the Basic Math Assistant palette.



I am new to vector calculus operations. There is a known identity found in my textbook.



$$qquad int _{4 pi }hat{s} (hat{s}cdot A) d omega=frac{4 pi}{3}A$$



I have no idea how to do this type of integration. This is what I tried, but it returns a disaster:



Integrate[s*(Dot[s, A]), s, {0, 4 π}]


Also without success:



Integrate[{Sin[θ], Cos[θ]}*(Dot[{Sin[θ], Cos[θ]}, {a1, a2}]), θ, {0, 4 π}]


It is obvious that I am doing something fundamentally not correct. I go to the documentation on Vector Calculus, but it does not offer much in substance or examples. How do you enter the integral expression shown above in order to return the identity in the right?



Update



In response to comments, here is a copy of the text. This is from page 10 of Optical-Thermal Response of Laser-Irradiated Tissue.



$omega$ is the surface area of a sphere in steradians. $hat s$ is the directional vector of a pencil of radiation located inside the sphere



table






symbolic vector-calculus




symbolic vector-calculus






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 11 mins ago









J. M. is slightly pensive♦

98.8k10311467




98.8k10311467










asked 3 hours ago









Jose Enrique CalderonJose Enrique Calderon

1,063718




1,063718












  • $begingroup$
    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♦
    3 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
    2 hours ago








  • 1




    $begingroup$
    I've never seen this author's notation. My guess is that $int_{4pi}cdots$ means the integral over the sphere of measure $4pi$, i.e., the unit sphere.
    $endgroup$
    – Michael E2
    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♦
    3 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
    2 hours ago








  • 1




    $begingroup$
    I've never seen this author's notation. My guess is that $int_{4pi}cdots$ means the integral over the sphere of measure $4pi$, i.e., the unit sphere.
    $endgroup$
    – Michael E2
    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♦
3 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♦
3 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
2 hours ago






$begingroup$
@Michael E2 please post it as an answear for upvote
$endgroup$
– Jose Enrique Calderon
2 hours ago






1




1




$begingroup$
I've never seen this author's notation. My guess is that $int_{4pi}cdots$ means the integral over the sphere of measure $4pi$, i.e., the unit sphere.
$endgroup$
– Michael E2
2 hours ago






$begingroup$
I've never seen this author's notation. My guess is that $int_{4pi}cdots$ means the integral over the sphere of measure $4pi$, i.e., the unit sphere.
$endgroup$
– Michael E2
2 hours ago












1 Answer
1






active

oldest

votes


















3












$begingroup$

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
    2 hours ago






  • 2




    $begingroup$
    @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♦
    2 hours 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
    2 hours 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, $hat{s}$ is implied to be a vector.
    $endgroup$
    – J. M. is slightly pensive♦
    2 hours 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









3












$begingroup$

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
    2 hours ago






  • 2




    $begingroup$
    @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♦
    2 hours 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
    2 hours 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, $hat{s}$ is implied to be a vector.
    $endgroup$
    – J. M. is slightly pensive♦
    2 hours 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
















3












$begingroup$

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
    2 hours ago






  • 2




    $begingroup$
    @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♦
    2 hours 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
    2 hours 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, $hat{s}$ is implied to be a vector.
    $endgroup$
    – J. M. is slightly pensive♦
    2 hours 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














3












3








3





$begingroup$

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 2 hours ago









Michael E2Michael E2

150k12203482




150k12203482












  • $begingroup$
    Why it simply does not work with limits of integration {s,0,4Pi}
    $endgroup$
    – Jose Enrique Calderon
    2 hours ago






  • 2




    $begingroup$
    @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♦
    2 hours 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
    2 hours 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, $hat{s}$ is implied to be a vector.
    $endgroup$
    – J. M. is slightly pensive♦
    2 hours 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
    2 hours ago






  • 2




    $begingroup$
    @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♦
    2 hours 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
    2 hours 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, $hat{s}$ is implied to be a vector.
    $endgroup$
    – J. M. is slightly pensive♦
    2 hours 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
2 hours ago




$begingroup$
Why it simply does not work with limits of integration {s,0,4Pi}
$endgroup$
– Jose Enrique Calderon
2 hours ago




2




2




$begingroup$
@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♦
2 hours 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.
$endgroup$
– J. M. is slightly pensive♦
2 hours 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
2 hours 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
2 hours 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, $hat{s}$ is implied to be a vector.
$endgroup$
– J. M. is slightly pensive♦
2 hours 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, $hat{s}$ is implied to be a vector.
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– J. M. is slightly pensive♦
2 hours ago






1




1




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@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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