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I got an assignment to prove that in a straight homogeneous rod, you can always choose a coordinate system in such a way that

$$\int_S x_1 \, dx_1 \, dx_2=0 $$

$$\int_S x_2 \, dx_1 \, dx_2=0 $$

$$\int_S x_1x_2 \, dx_1 \, dx_2=0$$

where $S$ is the cross section of the rod

Now, intuitively, I can understand that the first two expressions basically state that if you find the mass center of the rod, then all you have to do is to put the origin of the coordinate system in the mass center, and the integrals turn out to be zero. What I'm curious about is whether or not this is a good enough of an explanation, and also, a hint on how to obtain the third equality.

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  • $\begingroup$ Are all these supposed to be equal to zero? How are the first two different? Should the integrand be $x_2$? $\endgroup$
    – Ross Millikan
    Jan 14, 2014 at 23:54
  • $\begingroup$ oh, I'm sorry, one of them is supposed to be $\int_S x_2 dx_1dx_2$ and yes they are supposed to be zero $\endgroup$
    – user121399
    Jan 14, 2014 at 23:58
  • $\begingroup$ You should be able to edit that into the question. There should be an edit button just below the question. I don't think there is a rep minimum to edit your own question. If you don't see it, let me know and I will put it in $\endgroup$
    – Ross Millikan
    Jan 15, 2014 at 0:03

1 Answer 1

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For the third, put the origin at the centroid of the cross section of the rod. As you say, that will make the first two integrals zero. Pick the orientation of the $x_1$ axis randomly. Now imagine rotating the coordinate system $\frac \pi 2$ counterclockwise, plotting the integral as a function of rotation angle. It is clearly a continuous function. After the rotation, you can also get the same integral with the substitution $x_1 \to x_2, x_2 \to -x_1$, so the sign will invert. Somewhere in the rotation there must be a point where the value is zero.

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  • $\begingroup$ nice, thanks a lot $\endgroup$
    – user121399
    Jan 15, 2014 at 0:05

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