Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

15.3: Moment and Center of Mass

  • Page ID
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)

    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

    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:

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


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


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

    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


    \[\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

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


    • Was this article helpful?