My goal is to determine the moment of inertia of a square with the side length $a$. I know I could do this like this:

$$J = \frac{m}{a^2}\int_{\frac{-a}{2}}^{\frac{a}{2}} \int_{\frac{-a}{2}}^{\frac{a}{2}} x^2 \,dxdy = \frac{ma^2}{12}$$

But I thought that it would also work if I integrated differently over the square area. Let's define $r = \frac{a}{2}$ so $r$ has the same distance to every side. Now I can calculate the area of the square as follows:

$$\int_{0}^{\frac{a}2} 8r\, dr = a^2$$

And with this I tried to calculate the moment of inertia:

$$\int_{0}^{\frac{a}2} r^2\cdot8r\,dr = \frac{ma^2}{8}$$

It is easy to see that the result cannot be correct. But I don't understand why? I thought that area integration works over the perimeter for all polygons, isn't that correct?

  • $\begingroup$ $ \int_0^{\frac a2}8r\,\text dr $ is not the area of the square. $\endgroup$ – klunkean Jul 19 at 7:59
  • $\begingroup$ @klunkean why not? it is equal to $a^2$ which is the area of the square $\endgroup$ – Johny Dow Jul 19 at 8:03

Essentially you are using polar coordinates $r,\varphi$ with the origin in the center of the square. In this case $\text d r$ is a line segment along the radial coordinate and $r\text d \varphi$ would be an arclength segment on the perimeter of the circle with radius $r$.

What you calculate in your integral $$ \int_0^{\frac a2}8r\text d r $$ is $\frac{8}{2\pi}$ times the area of a circle with radius $\frac a2$ since

$$ A_\text{circle}=\int_{0}^{2\pi}\int_{0}^{\frac a2}r\text dr\text d\varphi=2\pi\int_0^{\frac a2}r\text d r=\pi\left(\frac a2\right)^2 $$

  • $\begingroup$ Thanks for this. Your answer was very helpful, especially the part where you explain what I actually compute. $\endgroup$ – Johny Dow Jul 19 at 9:33

The correct integral for area in polar coordinates is $$A =4\left( \int_{\phi = 0}^{\frac\pi4}\int_{r=0}^{\frac{a}{2\cos\phi}} r\,drd\phi + \int_{\phi =\frac\pi4}^{\frac\pi4}\int_{r=0}^{\frac{a}{2\sin\phi}} r\,drd\phi\right) = a^2$$

integrating along the two triangles in the first quadrant, and multiplying by $4$ due to symmetry.

Similarly for $I$, the squared distance to the $y$-axis is $x^2 = r^2\cos^2\phi$ so we have

$$I = \frac{m}{a^2}\cdot4\left( \int_{\phi = 0}^{\frac\pi4}\int_{r=0}^{\frac{a}{2\cos\phi}} r^3\cos^2\phi\,drd\phi + \int_{\phi =\frac\pi4}^{\frac\pi2}\int_{r=0}^{\frac{a}{2\sin\phi}} r^3\cos^2\phi\,drd\phi\right) = \frac{ma^2}{12}$$


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