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Here is a sketch of the problem statement : A cube of edge length $l$ is placed in three dimensional space with one vertex at the origin ${(0,0,0)}$ and all the faces parallel to the (Cartesian) coordinate planes. The cube lies entirely in the first "octant". Calculate the solid angle subtended by the cube face parallel to $YZ$ plane at the point $(0,0,l)$. There are two such faces parallel to the $YZ$ plane. The face I am referring to is one which does not have any of its vertices at the origin.

I apologize for not providing a drawing of the situation. I have a decent idea of "solid angle" concept, but I am not very familiar with tricky definite integrals for area and volume calculations. I can intuitively (and crudely) approximate solid angle by elementary methods, but I just can not use integrals for exact solid angle calculation. Can anyone help ? I can project the cube face on a spherical surface centered at $(0,0,l)$ and use the projected area to calculate solid angle. But how to calculate this projected area? I am unable to setup some suitable expression for the infinitesimal projected area $dA$. I just want a hint to start with. Thanks in advance.

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Imagine $8$ such cubes cuddling at $(0,0,0)$. Together they have an outer surface of $6\cdot4=24$ unit squares, and each such square encompasses the same solid angle with respect to $(0,0,0)$. The solid angle per unit quare therefore is $${4\pi\over24}={\pi\over6}\ ,$$ which is the answer to your question.

In order to compute this solid angle via an integral I put $l=1$ and write the projection in the form $${\bf f}:\quad (x,y,1)\mapsto{1\over\sqrt{1+x^2+y^2}}(x,y,1)\qquad\bigl((x,y)\in Q:=[0,1]^2\bigr)\ .$$ Now use $x$ and $y$ as parameters when computing the area of the surface $S:={\bf f}(Q)$. Compute ${\bf f}_x$, ${\bf f}_y$, and then the surface element $${\rm d}\omega=|{\bf f}_x\times{\bf f}_y|={1\over(1+x^2+y^2)^{3/2}}\>{\rm d}(x,y)\ .$$ (Deriving this result "geometrically" would of course require less computation.) It follows that $${\rm area}(S)=\int_0^1\int_0^1 {1\over(1+x^2+y^2)^{3/2}}\>dx\>dy\ .$$ I didn't do this integral myself. Mathematica computed $$\int_0^1{1\over(1+x^2+y^2)^{3/2}}\>dy={1\over(1+x^2)\,\sqrt{2+x^2}}$$ ad then arrived at the final result ${\pi\over6}$, as expected.

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  • $\begingroup$ Thanks for such an amazing(+1) solution. Of course, the way You solved the problem requires minimal calculations, by making use of symmetry arguments, but what I am really interested in is the more "rigorous" approach : (say) using definite integrals to calculate projected area, or maybe some other approach. This over-complicates everything, but I really want to learn more about use of definite integrals in area and volume calculations. Thanks again. $\endgroup$ – user399078 Apr 12 '18 at 5:32
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You are asking for the solid angle at one vertex of a cube subtended by a face of the cube that does not include that vertex.

It does not matter which vertex you choose; the cube is symmetric with respect to the vertices and the answer will be the same. So you might as well compute the angle at the origin of a face not touching the origin; it's easier to talk about.

If you want the solid angle of one of the three faces that do not have a vertex at the origin, remember that there are three of them, that they cover the entire octant, and that the cube has a three-way symmetry around their common vertex. This gives you an exact result that is not crude at all.

For a method that actually sets up an integral, note that the solid angle is simply the area of the projection of the cube's face onto a unit sphere around the origin. Take the method described here to approximate the projection of an arbitrary small square onto that sphere. This gives you a formula, $$\frac{{\mathbf x} \cdot \hat{\mathbf n}}{\lVert \mathbf x\rVert^2}A,$$ in which $\mathbf x$ is the displacement vector from the center of the sphere to a point in a small square of area $A$ and $\hat{\mathbf n}$ is a unit vector perpendicular to that square. When the center of the unit sphere is the origin, the coordinates of $\mathbf x$ are simply the coordinates of the small square; if you must place the center of the sphere at $(0,0,l),$ you will have to subtract coordinates in order to express $\mathbf x.$

In the limit, as the size of the square goes to zero, and adding up all the projections of the individual small squares within the face of the cube, you can replace the small area $A$ with the "infinitesimal" $\mathrm dA$ and integrate the projected area over the face of the cube.

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  • $\begingroup$ There are two cube faces parallel to the $YZ$ plane : one face (say $A$) with vertices $(0,0,0) , (0,0,l), (0,l,0)$ and $(0,l,l)$, and the other face (call it $B$) with vertices $(l,0,0), (l,l,0), (l,l,l)$ and $(l,0,l)$. I am referring to face $B$ in my question. My question is how to calculate the solid angle subtended by face $B$ at the point $(0,0,l)$. $\endgroup$ – user399078 Apr 11 '18 at 6:33
  • $\begingroup$ When I first read (and answered) the question, I thought you were referring to a face at the point $(0,0,l)$ (that is, touching that point). After re-reading the question a couple of times, I think you meant to describe a solid angle at that point. I have edited the answer accordingly. $\endgroup$ – David K Apr 11 '18 at 6:41
  • $\begingroup$ I completely agree with the symmetry part. But can You please provide an expression which allows the solid angle calculation ? You wrote "This gives an exact answer....", but what is this "this" ? You did not describe any method there. And about the link that You provided, I can not find anything that can help, not because I am very smart, but because I am just a beginner. This comment may sound rude to You, but I am just asking for the "Mathematics and equations and expressions" part of the solution. I can project the face on a sphere, but I have no idea how to calculate that projected area. $\endgroup$ – user399078 Apr 11 '18 at 6:51
  • $\begingroup$ You asked for a hints. I tried to give hints. Perhaps someone else will give better ones. $\endgroup$ – David K Apr 11 '18 at 6:52
  • $\begingroup$ What I actually meant was just an idea or a starting Mathematical expression, because I can not solve a Mathematics problem unless I write down something. All I could do was thinking of projecting the area on a sphere. I could not progress after that. $\endgroup$ – user399078 Apr 11 '18 at 6:57

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