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 11.1.)

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 17.3.1 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 11.1.4; it may be helpful to compare the two solutions.

Example 17.3.2 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, $\ds\bar x = \bar y = {8\over5\pi}$.

Exercises 17.3

Ex 17.3.1 Find the center of mass of a two-dimensional plate that occupies the square $[0,1]\times[0,1]$ and has density function $xy$. (answer)

Ex 17.3.2 Find the center of mass of a two-dimensional plate that occupies the triangle $0\le x\le1$, $0\le y\le x$, and has density function $xy$. (answer)

Ex 17.3.3 Find the center of mass of a two-dimensional plate that occupies the upper unit semicircle centered at $(0,0)$ and has density function $y$. (answer)

Ex 17.3.4 Find the center of mass of a two-dimensional plate that occupies the upper unit semicircle centered at $(0,0)$ and has density function $x^2$. (answer)

Ex 17.3.5 Find the center of mass of a two-dimensional plate that occupies the triangle formed by $x=2$, $y=x$, and $y=2x$ and has density function $2x$. (answer)

Ex 17.3.6 Find the center of mass of a two-dimensional plate that occupies the triangle formed by $x=0$, $y=x$, and $2x+y=6$ and has density function $\ds x^2$. (answer)

Ex 17.3.7 Find the center of mass of a two-dimensional plate that occupies the region enclosed by the parabolas $x=y^2$, $y=x^2$ and has density function $\ds\sqrt{x}$. (answer)

Ex 17.3.8 Find the centroid of the area in the first quadrant bounded by $x^2-8y+4=0$, $x^2=4y$, and $x=0$. (Recall that the centroid is the center of mass when the density is 1 everywhere.) (answer)

Ex 17.3.9 Find the centroid of one loop of the three-leaf rose $r=\cos(3\theta)$. (Recall that the centroid is the center of mass when the density is 1 everywhere, and that the mass in this case is the same as the area, which was the subject of exercise 11 in section 17.2.) The computations of the integrals for the moments $M_x$ and $M_y$ are elementary but quite long; Sage can help. (answer)

Ex 17.3.10 Find the center of mass of a two dimensional object that occupies the region $0\le x\le \pi$, $0\le y\le \sin x$, with density $\sigma=1$. (answer)

Ex 17.3.11 A two-dimensional object has shape given by $r=1+\cos\theta$ and density $\sigma(r,\theta)=2+\cos\theta$. Set up the three integrals required to compute the center of mass. (answer)

Ex 17.3.12 A two-dimensional object has shape given by $r=\cos\theta$ and density $\sigma(r,\theta)=r+1$. Set up the three integrals required to compute the center of mass. (answer)

Ex 17.3.13 A two-dimensional object sits inside $r=1+\cos\theta$ and outside $r=\cos\theta$, and has density $1$ everywhere. Set up the integrals required to compute the center of mass. (answer)