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Find the centroid of the region under the curve $y={e}^{x}$ over the interval $1\le x\le 3$ (see the following figure).
To compute the centroid, we assume that the density function is constant and hence it cancels out:
Thus the centroid of the region is
Calculate the centroid of the region between the curves $y=x$ and $y=\sqrt{x}$ with uniform density in the interval $0\le x\le 1.$
${x}_{c}=\frac{{M}_{y}}{m}=\frac{1\text{/}15}{1\text{/}6}=\frac{2}{5}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{y}_{c}=\frac{{M}_{x}}{m}=\frac{1\text{/}12}{1\text{/}6}=\frac{1}{2}$
For a clear understanding of how to calculate moments of inertia using double integrals, we need to go back to the general definition in Section $6.6.$ The moment of inertia of a particle of mass $m$ about an axis is $m{r}^{2},$ where $r$ is the distance of the particle from the axis. We can see from [link] that the moment of inertia of the subrectangle ${R}_{ij}$ about the $x\text{-axis}$ is ${({y}_{ij}^{*})}^{2}\rho ({x}_{ij}^{*},{y}_{ij}^{*})\text{\Delta}A.$ Similarly, the moment of inertia of the subrectangle ${R}_{ij}$ about the $y\text{-axis}$ is ${({x}_{ij}^{*})}^{2}\rho ({x}_{ij}^{*},{y}_{ij}^{*})\text{\Delta}A.$ The moment of inertia is related to the rotation of the mass; specifically, it measures the tendency of the mass to resist a change in rotational motion about an axis.
The moment of inertia ${I}_{x}$ about the $x\text{-axis}$ for the region $R$ is the limit of the sum of moments of inertia of the regions ${R}_{ij}$ about the $x\text{-axis}\text{.}$ Hence
Similarly, the moment of inertia ${I}_{y}$ about the $y\text{-axis}$ for $R$ is the limit of the sum of moments of inertia of the regions ${R}_{ij}$ about the $y\text{-axis}\text{.}$ Hence
Sometimes, we need to find the moment of inertia of an object about the origin, which is known as the polar moment of inertia. We denote this by ${I}_{0}$ and obtain it by adding the moments of inertia ${I}_{x}$ and ${I}_{y}.$ Hence
All these expressions can be written in polar coordinates by substituting $x=r\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta ,$ $y=r\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta ,$ and $dA=r\phantom{\rule{0.2em}{0ex}}dr\phantom{\rule{0.2em}{0ex}}d\theta .$ For example, ${I}_{0}={\displaystyle \underset{R}{\iint}{r}^{2}\rho \left(r\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta ,r\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta \right)}dA.$
Use the triangular region $R$ with vertices $\left(0,0\right),\left(2,2\right),$ and $\left(2,0\right)$ and with density $\rho \left(x,y\right)=xy$ as in previous examples. Find the moments of inertia.
Using the expressions established above for the moments of inertia, we have
Again use the same region $R$ as above and the density function $\rho \left(x,y\right)=\sqrt{xy}.$ Find the moments of inertia.
${I}_{x}={\displaystyle \underset{x=0}{\overset{x=2}{\int}}\phantom{\rule{0.2em}{0ex}}{\displaystyle \underset{y=0}{\overset{y=x}{\int}}{y}^{2}\sqrt{xy}\phantom{\rule{0.2em}{0ex}}dy\phantom{\rule{0.2em}{0ex}}dx}}=\frac{64}{35}$ and ${I}_{y}={\displaystyle \underset{x=0}{\overset{x=2}{\int}}\phantom{\rule{0.2em}{0ex}}{\displaystyle \underset{y=0}{\overset{y=x}{\int}}{x}^{2}\sqrt{xy}\phantom{\rule{0.2em}{0ex}}dy\phantom{\rule{0.2em}{0ex}}dx}}=\frac{64}{35}.$ Also, ${I}_{0}={\displaystyle \underset{x=0}{\overset{x=2}{\int}}\phantom{\rule{0.2em}{0ex}}{\displaystyle \underset{y=0}{\overset{y=x}{\int}}\left({x}^{2}+{y}^{2}\right)\sqrt{xy}\phantom{\rule{0.2em}{0ex}}dy\phantom{\rule{0.2em}{0ex}}dx}}=\frac{128}{21}.$
As mentioned earlier, the moment of inertia of a particle of mass $m$ about an axis is $m{r}^{2}$ where $r$ is the distance of the particle from the axis, also known as the radius of gyration .
Hence the radii of gyration with respect to the $x\text{-axis,}$ the $y\text{-axis,}$ and the origin are
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