
# 14.6: Moments and Centers of Mass

Using a single integral we were able to compute the center of mass for a one-dimensional object with variable density, and a two dimensional object with constant density. With a double integral we can handle two dimensions and variable density.

Just as before, the coordinates of the center of mass are

$$\bar x={M_y\over M} \qquad \bar y={M_x\over M},$$

where $$M$$ is the total mass, $$M_y$$ is the moment around the $$y$$-axis, and $$M_x$$ is the moment around the $$x$$-axis. (You may want to review the concepts in section 9.6.)

The key to the computation, just as before, is the approximation of mass. In the two-dimensional case, we treat density $$\sigma$$ as mass per square area, so when density is constant, mass is $$(\hbox{density})(\hbox{area})$$. If we have a two-dimensional region with varying density given by $$\sigma(x,y)$$, and we divide the region into small subregions with area $$\Delta A$$, then the mass of one subregion is approximately $$\sigma(x_i,y_j)\Delta A$$, the total mass is approximately the sum of many of these, and as usual the sum turns into an integral in the limit:

$$M=\int_{x_0}^{x_1}\int_{y_0}^{y_1} \sigma(x,y)\,dy\,dx,$$

and similarly for computations in cylindrical coordinates. Then as before

\eqalign{ M_x &= \int_{x_0}^{x_1}\int_{y_0}^{y_1} y\sigma(x,y)\,dy\,dx\cr M_y &= \int_{x_0}^{x_1}\int_{y_0}^{y_1} x\sigma(x,y)\,dy\,dx.\cr }

Example $$\PageIndex{1}$$

Find the center of mass of a thin, uniform plate whose shape is the region between $$y=\cos x$$ and the $$x$$-axis between $$x=-\pi/2$$ and $$x=\pi/2$$. Since the density is constant, we may take $$\sigma(x,y)=1$$.

It is clear that $$\bar x=0$$, but for practice let's compute it anyway. First we compute the mass:

$$M=\int_{-\pi/2}^{\pi/2} \int_0^{\cos x} 1\,dy\,dx =\int_{-\pi/2}^{\pi/2} \cos x\,dx =\left.\sin x\right|_{-\pi/2}^{\pi/2}=2.$$

Next,

$$M_x=\int_{-\pi/2}^{\pi/2} \int_0^{\cos x} y\,dy\,dx =\int_{-\pi/2}^{\pi/2} {1\over2}\cos^2 x\,dx={\pi\over4}.$$

Finally,

$$M_y=\int_{-\pi/2}^{\pi/2} \int_0^{\cos x} x\,dy\,dx =\int_{-\pi/2}^{\pi/2} x\cos x\,dx=0.$$

So $$\bar x=0$$ as expected, and $$\bar y=\pi/4/2=\pi/8$$. This is the same problem as in example 9.6.4; it may be helpful to compare the two solutions.

Example $$\PageIndex{2}$$

Find the center of mass of a two-dimensional plate that occupies the quarter circle $$x^2+y^2\le1$$ in the first quadrant and has density $$k(x^2+y^2)$$. It seems clear that because of the symmetry of both the region and the density function (both are important!), $$\bar x=\bar y$$. We'll do both to check our work.

Jumping right in:

$$M=\int_0^1 \int_0^{\sqrt{1-x^2}} k(x^2+y^2)\,dy\,dx =k\int_0^1 x^2\sqrt{1-x^2}+{(1-x^2)^{3/2}\over3}\,dx.$$

This integral is something we can do, but it's a bit unpleasant. Since everything in sight is related to a circle, let's back up and try polar coordinates. Then $$x^2+y^2=r^2$$ and

$$M=\int_0^{\pi/2} \int_0^{1} k(r^2)\,r\,dr\,d\theta =k\int_0^{\pi/2}\left.{r^4\over4}\right|_0^1\,d\theta =k\int_0^{\pi/2} {1\over4}\,d\theta =k{\pi\over8}.$$

Much better. Next, since $$y=r\sin\theta$$,

$$M_x=k\int_0^{\pi/2} \int_0^{1} r^4\sin\theta\,dr\,d\theta =k\int_0^{\pi/2} {1\over5}\sin\theta\,d\theta =k\left.-{1\over5}\cos\theta\right|_0^{\pi/2}={k\over5}.$$

Similarly,

$$M_y=k\int_0^{\pi/2} \int_0^{1} r^4\cos\theta\,dr\,d\theta =k\int_0^{\pi/2} {1\over5}\cos\theta\,d\theta =k\left.{1\over5}\sin\theta\right|_0^{\pi/2}={k\over5}.$$

Finally, $$\bar x = \bar y = {8\over5\pi}$$.

### Contributors

• Integrated by Justin Marshall.