<s>[INST] Why two different results to surface areal calculation using Guldin? Guldin's first rule, also known as Pappus first centroid theorem states that a linear figure creates an area when rotated that is the product of the distance the centroid is moved and the length of the figure.
I should therefore get the same resulting area when I rotate a triangle, as when I rotate the three segments making up the sides of the triangle.
Thus, I must do something wrong when I do not get that. Here are my calculations, provided triangle $ABC$, where $A=(1,1)$, $B=(1,4)$ og $C=(4,3)$.
In procedure 1, I shall calculate the contribution from each segment, then add.
In procedure 2, I shal find the centroid and the perimeter and "directly" calculate the surface area of the volume of rotation.
Procedure 1: 
$r_{AB}=1,L_{AB}=3,A_{AB}=2\cdot\pi\cdot r_{AB}\cdot L_{AB}=6\pi$.
$r_{BC}=\frac{5}{2},L_{BC}=\sqrt{(4-1)^2+(3-4)^2}=\sqrt{10},A_{BC}=2\cdot\pi\cdot r_{BC}\cdot L_{BC}=5\pi\sqrt{10}$.
$r_{AC}=\frac{5}{2},L_{AC}=\sqrt{(4-1)^2+(3-1)^2}=\sqrt{13},A_{AC}=2\cdot\pi\cdot r_{AC}\cdot L_{AC}=5\pi\sqrt{13}$.
Consequently, the total surface area amounts to $A_1=A_{AB}+A_{BC}+A_{AC}=\pi\left(6+5\sqrt{10}+5\sqrt{13}\right)$.
Procedure 2:
Triangle $ABC$ perimeter $L=L_{AB}+L_{BC}+L_{AC}=3+\sqrt{10}+\sqrt{13}$.
Radius of rotation for triangle $ABC$: $r=2$.
Total surface area: $A_2=2\pi r L=4\pi\left(3+\sqrt{10}+\sqrt{13}\right)$.
I expected $A_1=A_2$, but I get $A_1=A_2+\pi\left(-6+\sqrt{10}+\sqrt{13}\right)$.

 [/INST] The two plane figures below are not the same:

In partricular, their centroids are not in the same spot.
The centroid of the filled-in triangle at right is at (3, 3) as you found, but the centroid of the empty triangle at left is not.

The centroid of the empty triangle is at the weighted mean of the midpoints of the line segments.
Midpoints: $(1, 2.5), (2.5, 3.5), (2.5, 2)$
Weights: $3, \sqrt{10}, \sqrt{13}$
We only care about the $x$-coordinate of the centroid, though:
$\overline{x} = \frac{1 \cdot 3 + 2.5 \cdot \sqrt{10} + 2.5 \cdot \sqrt{13}}{3 + \sqrt{10} + \sqrt{13}}$
Multiply numerator and denominator by 2:
$\overline{x} = \frac{6 + 5 \sqrt{10} + 5 \sqrt{13}}{2\left(3 + \sqrt{10} + \sqrt{13}\right)}$

So the surface area calculated this way is $A_2 = 2 \pi r L = 2 \pi \overline{x} L$, which is

$A_2 = 2\pi\overline{x}L = 2\pi \frac{6 + 5 \sqrt{10} + 5 \sqrt{13}}{2\left(3 + \sqrt{10} + \sqrt{13}\right)} \left(3 + \sqrt{10} + \sqrt{13}\right)$
$A_2 = \pi \left(6 + 5 \sqrt{10} + 5 \sqrt{13}\right)$

, as desired.
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