# 15.3: Moment and Center of Mass

- Page ID
- 4823

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:

\[\begin{align*} M&=\int_{-\pi/2}^{\pi/2} \int_0^{\cos x} 1\,dy\,dx \\[4pt]&=\int_{-\pi/2}^{\pi/2} \cos x\,dx \\[4pt]&=\left.\sin x\right|_{-\pi/2}^{\pi/2}=2.\end{align*}\]

Next,

\[\begin{align*} M_x&=\int_{-\pi/2}^{\pi/2} \int_0^{\cos x} y\,dy\,dx \\[4pt]&=\int_{-\pi/2}^{\pi/2} {1\over2}\cos^2 x\,dx\\[4pt]&={\pi\over4}.\end{align*}\]

Finally,

\[\begin{align*} M_y&=\int_{-\pi/2}^{\pi/2} \int_0^{\cos x} x\,dy\,dx \\[4pt]&=\int_{-\pi/2}^{\pi/2} x\cos x\,dx\\[4pt]&=0.\end{align*}\]

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:

\[\begin{align*} M&=\int_0^1 \int_0^{\sqrt{1-x^2}} k(x^2+y^2)\,dy\,dx \\[4pt]&=k\int_0^1 x^2\sqrt{1-x^2}+{(1-x^2)^{3/2}\over3}\,dx. \end{align*}\]

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

\[\begin{align*} M&=\int_0^{\pi/2} \int_0^{1} k(r^2)\,r\,dr\,d\theta \\[4pt]&=k\int_0^{\pi/2}\left.{r^4\over4}\right|_0^1\,d\theta \\[4pt]&=k\int_0^{\pi/2} {1\over4}\,d\theta \\[4pt]&=k{\pi\over8}.\end{align*}\]

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

\[\begin{align*} M_x&=k\int_0^{\pi/2} \int_0^{1} r^4\sin\theta\,dr\,d\theta

\\[4pt]&=k\int_0^{\pi/2} {1\over5}\sin\theta\,d\theta

\\[4pt]&=k\left.-{1\over5}\cos\theta\right|_0^{\pi/2}={k\over5}.\end{align*}\]

Similarly,

\[\begin{align*} M_y&=k\int_0^{\pi/2} \int_0^{1} r^4\cos\theta\,dr\,d\theta

\\[4pt]&=k\int_0^{\pi/2} {1\over5}\cos\theta\,d\theta

\\[4pt]&=k\left.{1\over5}\sin\theta\right|_0^{\pi/2}={k\over5}.\end{align*}\]

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

## Contributors

Integrated by Justin Marshall.