# Why is this term 0 in the derivation of the parallel axis theorem?

I had a lab report to do based around moments of inertia and thought I'd remind myself of the parallel axis theorem, so I looked up the derivation. It goes something like:

$$I= \int [(x+D)^2 + y^2]dm ,$$

$$I = \int (x^2+y^2) dm + D^2\int dm + 2D\int xdm ,$$

where the last term is equal to zero. I was confused by this so I decided to look around a bit and none of the explanations have made sense to me. They explain that this term is equal to zero because the $$x$$ component of the center of mass is zero, if we have our origin as the center of mass. I can understand that, however, wouldn't that indicate that the x squared should also be equal to zero? I suppose the problem here is me not understanding exactly what these terms are saying.

$$\int xdm=0$$ by definition of center of mass, you are right. However, $$\int x^2 dm\neq 0$$ since it is a summation of positive (or rather non-negative) terms, $$x^2\geq 0$$.

Example: imagine two points with mass $$m$$ located at $$x=\pm1$$. Their center of mass is at $$x=0$$. Then $$\int xdm=-m+m=0$$, and $$\int x^2dm=m+m=2m$$.

• Alright, so it's because all of these are actually definite integrals, correct?. And therefore, as you said, the x^2 integral is positive. Thank you! This makes much more sense than the explanations I read. It's not really that the x-component is zero, rather the sum of all the mass-components to the right is equal to the sum of all the mass-components to the left. – agaminon Apr 2 at 16:48

The integral is similar to a sum, as $$dx$$ approaches zero.

$$\int x dm$$

does add up to zero at the centre of mass. If we take each $$dm$$ element as $$1$$, then it's similar to

$$1+2+3+(-1)+(-2)+(-3) = 0$$

(depending on what the $$x$$ and $$dm$$ weightings are, but the point is they cancel out).

$$\int x^2 dm$$

is different, it's simlar to adding like this $$1^2+2^2+3^2+(-1)^2+(-2)^2+(-3)^2 = 28$$