# Moment of inertia of a square (single integral)

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?

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

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

• Thanks for this. Your answer was very helpful, especially the part where you explain what I actually compute. – 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

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