Products Of Inertia And Rotated Axes

Why the Product of Inertia Matters

A lamina’s moments of inertia about the coordinate axes measure how its mass is distributed relative to each axis. But those two values do not describe every aspect of the distribution: mass in different quadrants can affect how the inertia changes when the axes are rotated. The product of inertia records this additional information.

For a lamina occupying a region $R$ in the $xy$-plane with density $\rho(x,y)$, the integrand $xy\rho(x,y)$ is positive in the first and third quadrants and negative in the second and fourth. Thus the product can be small or zero because of cancellation, even when the lamina has mass in all four quadrants. Its value depends on both the region and its density.

Definition: The product of inertia about the $x$- and $y$-axes is $$ P_{xy}=\iint_R xy\rho(x,y)\,dA. $$ The moments of inertia about those axes are $I_x=\iint_R y^2\rho(x,y)\,dA$ and $I_y=\iint_R x^2\rho(x,y)\,dA$. These moments and the product are taken about axes through the origin. If length is measured in units of length and density in units of mass per area, each has units of mass times length squared.

The sign convention here is important: the product is the integral of $xy\rho$, with no minus sign. The signs that appear in formulas for rotated axes follow from the coordinate transformation, not from changing this definition.

Symmetry and Vanishing Products

A reflection across either coordinate axis reverses the sign of $xy$. That fact gives a useful test for when the product is zero, provided the mass distribution—not just the outline of the region—has the corresponding symmetry.

Theorem: If $R$ is symmetric about the $x$-axis and the density satisfies $\rho(x,-y)=\rho(x,y)$, then $P_{xy}=0$. The same conclusion holds if $R$ is symmetric about the $y$-axis and $\rho(-x,y)=\rho(x,y)$.

For example, in the first case, the reflection $(x,y)\mapsto(x,-y)$ maps the region onto itself and leaves the density unchanged, but changes $xy\rho(x,y)$ to $-xy\rho(x,y)$. The contributions at reflected points cancel. The corresponding argument applies to reflection across the $y$-axis. A symmetric outline alone is not enough if the density fails to share that symmetry.

Worked Example: A Uniform Rectangle Centered at the Origin

Let $R=[-2,2]\times[-1,1]$ and let the constant density be $\rho(x,y)=3$. Compute $P_{xy}$ directly.

Since the region is rectangular, the integral separates into a product of one-dimensional integrals:

\begin{aligned} P_{xy} &=\int_{-2}^{2}\int_{-1}^{1}3xy\,dy\,dx\\ &=3\left(\int_{-2}^{2}x\,dx\right) \left(\int_{-1}^{1}y\,dy\right)\\ &=3(0)(0)=0. \end{aligned}

Each integral is zero because its integrand is odd over an interval symmetric about zero. This agrees with the reflection-symmetry test. The direct computation checks the result without relying on the test alone.

By contrast, symmetry through the origin—meaning that $(x,y)\in R$ exactly when $(-x,-y)\in R$—does not generally force $P_{xy}=0$. Under that transformation, the product $xy$ keeps its sign: $(-x)(-y)=xy$. The individual axis reflections are the relevant symmetries for this test.

Computing a Product on a Region Without Axis Symmetry

When no reflection symmetry forces cancellation, evaluate the defining double integral. The product can be positive or negative, depending on which quadrants contribute more strongly. Here is a region where it is positive.

Worked Example: A Triangular Lamina

Let $R$ be the triangle described by $0\leq y\leq x\leq1$, and let $\rho(x,y)=2$. Compute $P_{xy}$, along with the two moments needed to study a rotation.

Using $x$ as the outer variable and $y$ as the inner variable gives

\begin{aligned} P_{xy} &=2\int_0^1\int_0^x xy\,dy\,dx\\ &=2\int_0^1 x\left[\frac{y^2}{2}\right]_{0}^{x}\,dx\\ &=2\int_0^1 x\left(\frac{x^2}{2}\right)\,dx\\ &=\int_0^1x^3\,dx =\left[\frac{x^4}{4}\right]_0^1 =\frac14. \end{aligned}

For the moment about the $x$-axis, integrate $y^2\rho$:

\begin{aligned} I_x &=2\int_0^1\int_0^x y^2\,dy\,dx =2\int_0^1\frac{x^3}{3}\,dx =\frac{2}{3}\left[\frac{x^4}{4}\right]_0^1 =\frac16. \end{aligned}

For the moment about the $y$-axis, integrate $x^2\rho$:

\begin{aligned} I_y &=2\int_0^1\int_0^x x^2\,dy\,dx =2\int_0^1x^2(x-0)\,dx\\ &=2\int_0^1x^3\,dx =2\left[\frac{x^4}{4}\right]_0^1 =\frac12. \end{aligned}

All three values are positive, as expected here because $x$ and $y$ are nonnegative throughout the triangle. Their units are mass times length squared when the coordinates carry length units.

Rotating the Coordinate Axes

Now rotate the coordinate axes counterclockwise through an angle $\theta$, keeping the origin fixed. The coordinates of the same point in the rotated system are defined by

x'=x\cos\theta+y\sin\theta, \qquad y'=-x\sin\theta+y\cos\theta.

This convention is stated explicitly because reversing a coordinate transformation can reverse signs in the product formula. The rotated product is defined in the same way as before, using the new coordinates: $P_{x'y'}=\iint_R x'y'\rho\,dA$.

Theorem: For this counterclockwise rotation of the coordinate axes, $$ P_{x'y'}=P_{xy}\cos(2\theta)+\frac{I_x-I_y}{2}\sin(2\theta). $$ The rotated moments are $$ I_{x'}=I_x\cos^2\theta+I_y\sin^2\theta-P_{xy}\sin(2\theta), \qquad I_{y'}=I_x\sin^2\theta+I_y\cos^2\theta+P_{xy}\sin(2\theta). $$

These formulas apply to moments about the rotated axes through the same origin, for the same lamina and density. They do not silently shift the axes to the centroid. If the origin or the axes’ location changes, the moments must be computed for that new location.

Why the Rotated Product Formula Works

Substitute the coordinate expressions into $x'y'$, then integrate each term against the density. Writing $c=\cos\theta$ and $s=\sin\theta$, the product expands as follows:

\begin{aligned} x'y' &=(xc+ys)(-xs+yc)\\ &=-x^2cs+xyc^2-xys^2+y^2sc\\ &=-x^2cs+xy(c^2-s^2)+y^2sc. \end{aligned}

By the definitions of $I_x$, $I_y$, and $P_{xy}$, the integrals of $y^2\rho$, $x^2\rho$, and $xy\rho$ are $I_x$, $I_y$, and $P_{xy}$, respectively. Therefore, using $2sc=\sin(2\theta)$ and $c^2-s^2=\cos(2\theta)$,

\begin{aligned} P_{x'y'} &=-I_ycs+P_{xy}(c^2-s^2)+I_xsc\\ &=P_{xy}\cos(2\theta)+\frac{I_x-I_y}{2}\sin(2\theta). \end{aligned}

This derivation explains both the doubled angle and the sign of the moment-difference term. The same substitution, followed by squaring $y'$ or $x'$, gives the rotated-moment formulas stated above. As a consistency check, their sum is unchanged: $I_{x'}+I_{y'}=I_x+I_y$. The cross terms cancel, and $\cos^2\theta+\sin^2\theta=1$.

Using the Formula in Examples

Worked Example: Rotate the Triangular Lamina by 45 Degrees

For the triangular lamina above, $I_x=\frac16$, $I_y=\frac12$, and $P_{xy}=\frac14$. Set $\theta=45^\circ$, so $\cos(2\theta)=\cos90^\circ=0$ and $\sin(2\theta)=\sin90^\circ=1$. Then

\begin{aligned} P_{x'y'} &=\frac14(0)+\frac{\frac16-\frac12}{2}(1)\\ &=\frac{\frac16-\frac36}{2} =\frac{-\frac26}{2} =-\frac16. \end{aligned}

Check by expanding the rotated coordinates directly. At $45^\circ$, $x'y'=\frac12(y^2-x^2)$, so

\begin{aligned} P_{x'y'} &=\iint_R \frac{y^2-x^2}{2}\rho\,dA\\ &=\frac12(I_x-I_y) =\frac12\left(\frac16-\frac12\right) =-\frac16. \end{aligned}

The direct calculation agrees with the rotation formula, including its negative sign.

Worked Example: A Centered Rectangle with Unequal Moments

Take $R=[-1,1]\times[-\frac12,\frac12]$ with constant density $\rho=12$. Reflection symmetry gives $P_{xy}=0$. Compute the original moments and the product after a $45^\circ$ rotation.

The moment about the $x$-axis is

\begin{aligned} I_x &=12\int_{-1}^{1}\int_{-\frac12}^{\frac12}y^2\,dy\,dx\\ &=12(2)\left[\frac{y^3}{3}\right]_{-\frac12}^{\frac12} =24\left(\frac{1}{24}-\left(-\frac{1}{24}\right)\right) =2. \end{aligned}

The moment about the $y$-axis is

\begin{aligned} I_y &=12\int_{-1}^{1}\int_{-\frac12}^{\frac12}x^2\,dy\,dx\\ &=12(1)\left[\frac{x^3}{3}\right]_{-1}^{1} =12\left(\frac13-\left(-\frac13\right)\right) =8. \end{aligned}

At $45^\circ$, the formula gives

P_{x'y'}=0(0)+\frac{2-8}{2}(1)=-3.

Check directly: $x'y'=\frac12(y^2-x^2)$ at this angle, so the integral is $\frac12(I_x-I_y)=\frac12(2-8)=-3$, as required.

A Density Can Break the Symmetry Test

When checking symmetry, it is the weighted mass distribution that matters. A region may be symmetric about both coordinate axes while its density is not. The following example makes the distinction visible and also shows how rotation can make a nonzero product vanish.

Worked Example: Symmetric Square with Asymmetric Density

Let $R=[-1,1]\times[-1,1]$ and $\rho(x,y)=1+xy$. Since $-1\leq xy\leq1$, this density is nonnegative. The square is symmetric about each axis, but the density changes under reflection across just one axis. Compute the product and rotate by $45^\circ$.

First,

\begin{aligned} P_{xy} &=\int_{-1}^{1}\int_{-1}^{1}xy(1+xy)\,dy\,dx\\ &=\int_{-1}^{1}\int_{-1}^{1}(xy+x^2y^2)\,dy\,dx\\ &=0+\left(\int_{-1}^{1}x^2\,dx\right) \left(\int_{-1}^{1}y^2\,dy\right)\\ &=\frac23\cdot\frac23 =\frac49. \end{aligned}

The $xy$ term integrates to zero by oddness, while the $x^2y^2$ term contributes positively. Next,

\begin{aligned} I_x &=\iint_R y^2(1+xy)\,dA\\ &=\iint_R y^2\,dA+\iint_R xy^3\,dA =\frac43+0=\frac43, \\[4pt] I_y &=\iint_R x^2(1+xy)\,dA =\iint_R x^2\,dA+\iint_R x^3y\,dA =\frac43+0=\frac43. \end{aligned}

Since $I_x=I_y$, the rotated-product formula reduces to $P_{x'y'}=P_{xy}\cos(2\theta)$. At $\theta=45^\circ$, this gives $P_{x'y'}=\frac49(0)=0$. Directly, $x'y'=\frac12(y^2-x^2)$, whose integral is $\frac12(I_x-I_y)=0$. Thus the rotated value is verified. The original product is nonzero precisely because the density’s weighting distinguishes the quadrants.

Choosing Axes and Avoiding a Sign Error

The formula can help identify a rotation that makes the product zero. The condition is

P_{xy}\cos(2\theta)+\frac{I_x-I_y}{2}\sin(2\theta)=0.

When $I_x\ne I_y$, a suitable angle satisfies $\tan(2\theta)=\frac{2P_{xy}}{I_y-I_x}$, provided the angle is interpreted with the appropriate quadrant and the equation’s sine and cosine are both considered. If $I_x=I_y$ and $P_{xy}\ne0$, the condition instead requires $\cos(2\theta)=0$. If $I_x=I_y$ and $P_{xy}=0$, the product is zero for every rotation. These statements concern the product alone; they do not change the definition of the moments.

A frequent mistake is to write the difference term as $\frac{I_y-I_x}{2}\sin(2\theta)$ while using the coordinate transformation given here. For the rectangle in the preceding example, that incorrect sign would give $+\frac{8-2}{2}=+3$ at $45^\circ$. But direct expansion gives $\frac12(I_x-I_y)=\frac12(2-8)=-3$. The correction is to preserve the order $I_x-I_y$ in the formula. Expanding $x'y'$ before substituting moment values is a reliable way to check the sign.

The product of inertia matters because it captures how the mass distribution is oriented relative to the chosen axes. Reflection symmetry can eliminate it immediately, while a general distribution requires calculation. Once $I_x$, $I_y$, and $P_{xy}$ are known about a common origin, the rotation formulas show how all three quantities respond to a change of axis direction.

Check Your Understanding

Use the definitions and rotation convention in this tutorial to answer the following.

What reflection symmetries of the region and density guarantee that $P_{xy}=0$?
Why does symmetry under $(x,y)\mapsto(-x,-y)$ alone not force the product to vanish?
For the stated coordinate transformation, which order of subtraction appears in the rotated-product formula: $I_x-I_y$ or $I_y-I_x$?
If $I_x=I_y$, how does the formula for $P_{x'y'}$ simplify?
For the triangular lamina, what is $P_{x'y'}$ after a $45^\circ$ rotation, and what direct expression checks the result?