Parallel axis theorem for product of inertia

Statement

The product of inertia of an area about any pair of perpendicular axes $X$–$Y$ is equal to the product of inertia about the parallel centroidal axes plus the product of the area and the coordinates of the centroid with respect to the $X$–$Y$ axes:

$$I_{XY} = I_{xy} + A\,\bar x\,\bar y$$

where $I_{xy}$ is the product of inertia about the centroidal axes $x$, $y$ (parallel to $X$, $Y$), and $(\bar x, \bar y)$ are the coordinates of the centroid G in the $X$–$Y$ system, with signs.

Proof

Let $dA$ be an element with coordinates $(x, y)$ with respect to the centroidal axes. Its coordinates with respect to the $X$–$Y$ axes are

$$X = x + \bar x,\qquad Y = y + \bar y$$

Then

$$\begin{aligned} I_{XY} &= \int XY\,dA = \int (x + \bar x)(y + \bar y)\,dA\\ &= \int xy\,dA + \bar y\int x\,dA + \bar x\int y\,dA + \bar x\bar y\int dA \end{aligned}$$

Since $x$ and $y$ are measured from the centroid, $\int x\,dA = 0$ and $\int y\,dA = 0$. Also $\int xy\,dA = I_{xy}$ and $\int dA = A$. Therefore

$$I_{XY} = I_{xy} + A\bar x\bar y\qquad\blacksquare$$

For a composite section the same formula is applied to each part and the results are added: $I_{xy} = \sum (\bar I_{xy,i} + A_i d_{x,i}d_{y,i})$.