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At present, I teach Guldin's rules for surface and volume of rotation, and give an example task from the textbook. The textbook uses procedure 1 (below) for calculation (below), but I advocate that you should be able to use procedure 2 as well. Where is the flaw in my argument for equivalence between the two procedures (or in my calculation of the surface area)?

Guldin's first rule is also known as Pappus first centroid theorem, and states that a plane figure creates an areal 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 $A_1=A_2+\pi\left(-6+\sqrt{10}+\sqrt{13}\right)$. Calculation of surface area of triangle rotated around the y-axis

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    $\begingroup$ This question is really a mathematical one, and not about pedagogy/education. Better fits on Mathematics SE. Vote to close. $\endgroup$ – Daniel R. Collins Apr 22 '18 at 16:37
  • $\begingroup$ It's so tempting to give the answer though :D $\endgroup$ – Chris Cunningham Apr 23 '18 at 11:31
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    $\begingroup$ @ChrisCunningham you can do so over there math.stackexchange.com/questions/2744680/… $\endgroup$ – quid Apr 24 '18 at 10:32
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    $\begingroup$ Ha -- fair point. Interesting that I never considered it. Thanks @quid $\endgroup$ – Chris Cunningham Apr 24 '18 at 14:52
  • $\begingroup$ Answer in Mathematics by @ChrisCunningham was exactly what I was looking for: Area centroid may differ from periphery (linear) centroid. $\endgroup$ – Engelsmann Apr 24 '18 at 17:26