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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

UPDATE 2

Will like to share that a proof to this was included in a Mathematic post here

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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.'s missing motivation ♦
    Commented Apr 1, 2019 at 1:23
  • 3
    $\begingroup$ 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
    Commented Apr 1, 2019 at 1:33
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    $\begingroup$ Also to stave off possible questions about entry #5: Integrate[{x, y, z}, {x, y, z} ∈ RegionIntersection[Sphere[], HalfSpace[{0, 0, -1}, 0]]]. Entry #6 would instead use HalfSpace[{0, 0, 1}, 0]. $\endgroup$
    – J. M.'s missing motivation ♦
    Commented Apr 1, 2019 at 2:24
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    $\begingroup$ I've never seen this author's notation. My guess is that $\int_{4\pi}\cdots$ means the integral over the sphere of measure $4\pi$, i.e., the unit sphere. $\endgroup$
    – Michael E2
    Commented Apr 1, 2019 at 2:27
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    $\begingroup$ I think the key point here is to understand the phrase solid angle: mathworld.wolfram.com/SolidAngle.html $\endgroup$
    – xzczd ♦
    Commented Apr 25, 2019 at 13:03

1 Answer 1

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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  *)
Improve this answer
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  • $\begingroup$ Why it simply does not work with limits of integration {s,0,4Pi} $\endgroup$
    – Jose Enrique Calderon
    Commented Apr 1, 2019 at 2:15
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    $\begingroup$ @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.'s missing motivation ♦
    Commented Apr 1, 2019 at 2:17
  • $\begingroup$ @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
    Commented Apr 1, 2019 at 2:20
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    $\begingroup$ @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.'s missing motivation ♦
    Commented Apr 1, 2019 at 2:23
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    $\begingroup$ @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.'s missing motivation ♦
    Commented Apr 1, 2019 at 2:31

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