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Being radiation a vector, the resolution for a third coordinate plane that obviously cuts the emitting circle in two halves is required; the outline of the integral in this case yields:
In this particular case, the limits of the integral cannot be extended to 2n, as the value would be nil. If (41) is integrated with respect to 0, in the numerator the derivatives of cos0, – sin0 could be found. Therefore by making this change:
t = cosQ dt = -sin0d0 (42)
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Integral (41) can therefore be expressed as:
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Taking out all the constants, and integrating with respect to r, the primitive is just the quotient of the numerator:
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Integrating (44) with respect to r; the last integral to solve is:
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That responds to the form:
xdx 1 7 T_ h dx T_ 2 I = 4nX – I X = ax + hx + c D = 4ac – h
J X 2a 2aJ X
dx 2 2ax + b
= – arctan
3 X D D
Substituting in (46) yields for both terms:
i = 1; b = ±2d; c = ( d2 + b1)
X = r2 ±2dr + (d2 + b2) D = 4b2
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Now, substituting in the first term of (45):
And bringing equation (48) to the limits of the integral [a,0]:
2(ln(a2 + d2 + b2 -2da))-ln(d2 + b2) + b(arctana ^ + arctanb
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In a similar way, for the second term of equation (45):
Finally, multiplying by b and grouping, the final result can be expressed as:
1 ( a + d a – d Л b a2 + b2 + d2 – 2ad
— I arctan——– + arctan—— 1+——– in——- 22————
2pl b b 0 4pd a2 + b2 + d2 + 2ad
It can be demonstrated that the former equates the area subtended by a circular sector that encompasses the diameter of the emitting disk and the corresponding sector of a hyperbola defined by the intersection of the unit sphere and the cone [7].
If a=d, the factor is,
1 2a b, 4a2 + b
—arctan————– in—- 2—
2p b 4pd b2
Where the former assert is more easily visualized.
If d=0 the expression is undetermined and the limit is passed with l’Hopital’s rule, obtaining the familiar result.
a
arctan – b
Equation (51) is entirely new and has never been mentioned in literatures; equation (52) constitutes a particular case of the former, that is, when a equals d, meaning that the receiving point lies on the edge of the emitting semicircle; equation (53) is the particular case in which the receiving point is aligned with the center of the half-disk.
