## 3 Results: Solutions for Various Forms of Triangles and Trapezes
### Right Triangle with Minor Vertex at the Origin of Coordinates
\[f_{21}=\frac{1}{2\pi}\left(arctan\frac{b-x}{y}+arctan\frac{x}{y}-\frac{y}{\sqrt{a^ {2}+y^{2}}}\left(arctan\frac{b-x}{\sqrt{a^{2}+y^{2}}}+arctan\frac{x}{\sqrt{a^{2}+ y^{2}}}\right)\right) \tag{18}\]  
Let us now calculate, for the sake of generality, the same triangle in a more general position outside the origin of coordinates, see Figure 4.  
In this case, integration limits in z, would change between,  
\[z=\frac{a}{b}x-\frac{am}{b}\text{and z = 0}\] \[\int_{0}^{\frac{x}{x-\frac{am}{b}}}\frac{yzdz}{\left[(x-x_{0})^{2 }+y^{2}+z^{2}\right]^{2}}=-\frac{y}{2}\left[\frac{1}{(x-x_{0})^{2}+y^{2}+z^{2} }\right]_{0}^{\frac{x}{x-\frac{am}{b}}} \tag{19}\] \[=-\frac{y}{2}\left[\frac{1}{(x-x_{0})^{2}+y^{2}+z^{2}}\right]_{0} ^{\frac{x}{x-\frac{am}{b}}}=\frac{y}{2}\left(\frac{1}{(x-x_{0})^{2}+y^{2}}- \frac{1}{(x-x_{0})^{2}+y^{2}+\left(\frac{x}{x-\frac{am}{b}}\right)^{2}}\right)\]  
But  
\[\left(\frac{a}{b}x-\frac{am}{b}\right)^{2}=\frac{a^{2}}{b^{2}}x^{2}-2\frac{a^{ 2}}{b^{2}}mx-\frac{a^{2}m^{2}}{b^{2}}=\frac{a^{2}}{b^{2}}\left(x^{2}-2 mx+m^{2}\right) \tag{20}\]  
And regrouping, the partial result is  
\[\frac{y}{2}\left(\frac{1}{(x-x_{0})^{2}+y^{2}}-\frac{1}{(x-x_{0})^{2}+y^{2}+ \frac{a^{2}(x^{2}-2 mx+m^{2})}{b^{2}}}\right) \tag{21}\]  
Conveniently developed, this gives,  
\[\frac{y}{2}\left(\frac{1}{(x-x_{0})^{2}+y^{2}}-\frac{b^{2}}{b^{2}(x-x_{0})^{2 }+b^{2}y^{2}+a^{2}x^{2}-2a^{2}mx+a^{2}m^{2}}\right) \tag{22}\]  
That is,  
\[\frac{y}{2}\left(\frac{1}{(x-x_{0})^{2}+y^{2}}-\frac{b^{2}}{b^{2}(x-x_{0})^{2 }+b^{2}y^{2}+a^{2}(x^{2}-2 mx+m^{2})}\right) \tag{23}\]  
Figure 4: Coordinates for the triangle form factor over a horizontal plane. The first term of the integral is already solved in Equation (10),  
Figure 4.0  
For the second part, we need to integrate,  
Figure 4.1  
Developing all the terms in the denominator, we receive,  
\[ \begin{split} b^{2}\left(x-x_{0}\right)^{2}+b^{2}y^{2}+a^{2} \left(x^{2}-2 mx+m^{2}\right)\\ =b^{2}\left(x^{2}+x_{0}{}^{2}-2 xx_{0}\right)+b^{2}y^{2}+a^{2}x^{2 }-2a^{2}mx+a^{2}m^{2}\end{split}  \tag{26}\]  
And grouping in terms of x,  
\[\left(a^{2}+b^{2}\right)x^{2}-2\left(b^{2}x_{0}+a^{2}m\right)x+b^{2}\left(y^{ 2}+x_{0}{}^{2}\right)+a^{2}m^{2} \tag{27}\]  
The final integral to be solved is,  
\[ \begin{split}&\int_{m}^{b+m}\frac{b^{2}dx}{\left(a^{2}+b^{2} \right)x^{2}-2\left(b^{2}x_{0}+a^{2}m\right)x+b^{2}\left(y^{2}+x_{0}{}^{2} \right)+a^{2}m^{2}}=\\ &=\frac{2b^{2}}{b\sqrt{a^{2}\left(y^{2}+\left(x_{0}-m\right)^{2} \right)+b^{2}y^{2}}}\bigg{[}arctan\frac{2x\left(a^{2}+b^{2}\right)-2\left(a^{ 2}m+b^{2}x_{0}\right)}{2b\sqrt{a^{2}\left(x_{0}-m\right)^{2}+b^{2}y^{2}}} \bigg{]}^{b+m}\\ &=\frac{2b}{\sqrt{a^{2}\left(x_{0}-m\right)^{2}+y^{2}\left(a^{ 2}+b^{2}\right)}}\bigg{[}arctan\frac{b\left(a^{2}+b\left(b+m-x_{0}\right) \right)}{b\sqrt{a^{2}\left(x_{0}-m\right)^{2}+y^{2}\left(b^{2}+a^{2}\right)}} \\ &-arctan\frac{b^{2}\left(m-x_{0}\right)}{b\sqrt{a^{2}\left(x_{0} -m\right)^{2}+y^{2}\left(b^{2}+a^{2}\right)}}\bigg{]}\end{split}  \tag{28}\]  
\[ \begin{split}&=\frac{2b}{\sqrt{a^{2}\left(x_{0}-m\right)^{2}+y^{2} \left(a^{2}+b^{2}\right)}}\bigg{[}arctan\frac{a^{2}+b\left(b+m-x_{0}\right) }{\sqrt{a^{2}\left(x_{0}-m\right)^{2}+y^{2}\left(a^{2}+b^{2}\right)}}\\ &-arctan\frac{b\left(m-x_{0}\right)}{\sqrt{a^{2}\left(x_{0}-m \right)^{2}+y^{2}\left(a^{2}+b^{2}\right)}}\bigg{]}\end{split}  \tag{29}\]  
Arranging all the necessary terms and multiplying by y/2, we obtain,  
\[ \begin{split}& f_{21}\\ &=\frac{1}{2\pi}\Big{[}arctan\frac{b+m-x_{0}}{y}+arctan\frac{x_ {0}-m}{y}\Big{]}\\ &-\frac{by}{\sqrt{a^{2}\left(x_{0}-m\right)^{2}+y^{2}\left(a^{ 2}+b^{2}\right)}}\Big{[}arctan\frac{a^{2}+b\left(b+m-x_{0}\right)}{\sqrt{a^{2 }\left(x_{0}-m\right)^{2}+y^{2}\left(a^{2}+b^{2}\right)}}\\ &-arctan\frac{b\left(m-x_{0}\right)}{\sqrt{a^{2}\left(x_{0}-m \right)^{2}+y^{2}\left(a^{2}+b^{2}\right)}}\Big{]}\end{split}  \tag{30}\]  
By making m = 0, we would find that,  
\[