Multiple integrals: Applications of multiple integrals
Centre of gravity, static moments and moments of inertia
If the continuous density \(\rho(x,y,z)\) is given in every point of a rigid body \(R\), then for a volume element with very small dimensions \(\dd (x,y,z)\) it is true that its mass is approximately equal to \(\rho(x,y,z)\,\dd(x, y,z)\). The total mass \(M\) of \(R\) is therefore given by \[M=\iiint_R \rho(x,y,z)\,\dd(x, y,z)\]
Static moments in three dimensions The 3D static moments \(M_{yz}\), \(M_{xz}\) and \(M_{xy}\) of \(R\) relative to the \(yz\)-plane, the \(xz\)-plane and the \(xy\)-plane, respectively, also called first (mass)moments, are defined as follows:
\[\begin{aligned} M_{yz} &=\iiint_R x\,\rho(x,y,z)\,\dd(x,y,z)\\[0.25cm]
M_{xz} &=\iiint_R y\,\rho(x,y,z)\,\dd(x,y,z)\\[0.25cm]
M_{xy} &=\iiint_R z\,\rho(x,y,z)\,\dd(x,y,z)\end{aligned}\]
The 3D centre of gravity, usually called the centroid, \((\bar{x},\bar{y},\bar{z})\) of \(R\) is given by: \[\bar{x}=\frac{M_{yz}}{M},\quad \bar{y}=\frac{M_{xz}}{M},\quad \bar{z}=\frac{M_{xy}}{M}\]
Determine the centre of gravity of the rigid body \(R\) sketched below and enclosed by the plane \(z=0\) and the paraboloid \(z=4-x^2-y^2\) under the assumption that the density is a constant \(\rho\).
\(\cv{\bar{x} \\ \bar{y} \\ \bar{z}} ={}\) \(\cv{0\\ 0\\ {{4}\over{3}}}\)
By symmetry, we know that \(\bar{x}=0\) and \(\bar{y}=0\). We can therefore limit ourselves to the calculation of \(\bar{z}\).
First we calculate the total mass \(M\) : \[\begin{aligned} M &= \iiint_R \rho\,\dd(x,y,z)\\[0.25cm] &= \rho\, \int_{z=0}^{z=4}\left(\int_{\varphi=0}^{\varphi=2\pi}\left(\int_{r=0}^{r=\sqrt{4-z}}r\,\dd r\right)\dd\varphi\right)\dd z\\[0.25cm]&\phantom{abcuva} \blue{\text{in cylindrical coordinates}}\\[0.25cm] &= \rho\, \int_{z=0}^{z=4}\left(\int_{\varphi=0}^{\varphi=2\pi}\biggl[\frac{1}{2}r^2\biggl]_{r=0}^{r=\sqrt{4-z}}\;\dd\varphi\right)\dd z\\[0.25cm] &=\frac{1}{2}\rho\int_{z=0}^{z=4}\left(\int_{\varphi=0}^{\varphi=2\pi}(4-z)\,\dd \varphi\right)\dd z\\[0.25cm] &=\frac{1}{2}\rho\int_{z=0}^{z=4}\biggl[(4-z)\,\varphi\biggr]_{\varphi=0}^{\varphi=2\pi}\;\dd z \\[0.25cm] &= \pi\,\rho\int_{z=0}^{z=4}(4-z)\,\dd z\\[0.25cm] &= \pi\,\rho\,\biggl[-\frac{1}{2}(4-z)^2\biggr]_{z=0}^{z=4}\\[0.25cm] &=\pi\,\rho\, \left(0-\Bigl(-\frac{1}{2}\,4^2\Bigr)\right)\\[0.25cm] &= 8\,\pi\,\rho\end{aligned}\] Then we determine the static moment in a similar way \(M_{xy}\) : \[\begin{aligned} M_{xy} &= \iiint_R z\,\rho\,\dd(x,y,z)\\[0.25cm] &= \rho\, \int_{z=0}^{z=4}\left(\int_{\varphi=0}^{\varphi=2\pi}\left(\int_{r=0}^{r=\sqrt{4-z}}z\,r\,\dd r\right)\dd\varphi\right)\dd z\\[0.25cm]&\phantom{abcuva} \blue{\text{in cylindrical coordinates}}\\[0.25cm] &= \rho\, \int_{z=0}^{z=4}\left(\int_{\varphi=0}^{\varphi=2\pi}\biggl[\frac{1}{2}r^2\,z\biggl]_{r=0}^{r=\sqrt{4-z}}\;\dd\varphi\right)\dd z\\[0.25cm] &=\frac{1}{2}\rho\int_{z=0}^{z=4}\left(\int_{\varphi=0}^{\varphi=2\pi}(4-z)\,z\,\dd \varphi\right)\dd z\\[0.25cm] &=\frac{1}{2}\rho\int_{z=0}^{z=4}\biggl[(4-z)\,z\,\varphi\biggr]_{\varphi=0}^{\varphi=2\pi}\;\dd z \\[0.25cm] &= \pi\,\rho\int_{z=0}^{z=4}(4-z)\,z\,\dd z\\[0.25cm] &= \pi\,\rho\,\biggl[2z^2-\frac{1}{3}z^3\biggr]_{z=0}^{z=4}\\[0.25cm] &=\pi\,\rho\, \left(\frac{1}{2}\,4^2-\frac{1}{3}\,4^3-0\right)\\[0.25cm] &= {{32}\over{3}}\,\pi\,\rho\end{aligned}\] The \(z\)-coordinate of the centre of gravity can now be calculated as a quotient: \[\begin{aligned} \bar{z}&=\frac{M_{xy}}{M}\\[0.25cm] &= \frac{{{32}\over{3}}\,\pi\,\rho}{8\,\pi\,\rho}\\[0.25cm] &= {{4}\over{3}}\end{aligned}\]
Moments of inertia in three dimensions The 3D moment of inertia \(M_{\ell}\) of \(R\) relative to the line \(\ell\) is defined as follows: \[I_{\ell} = \iiint \delta(x,y,z)^2\,\rho(x,y,z)\,\dd(x,y,z)\] where \[\delta(x,y,z)^2=\text{distance from the point }(x,y,z)\text{ to the line }\ell\text.\] Three special cases are the moments of inertia relative to the coordinate axes: \[\begin{aligned} I_x &=\iiint_R (y^2+z^2)\,\rho(x,y,z)\,\dd(x,y,z)\\[0.25cm]
I_{y} &=\iiint_R (x^2+z^2) \,\rho(x,y,z)\,\dd(x,y,z)\\[0.25cm]
I_{z} &=\iiint_R (x^2+y^2)\,\rho(x,y,z)\,\dd(x,y,z)\end{aligned}\] The radius of inertia \(k_{\ell}\) relative to a line \(\ell\) is defined in terms of the moment of inertia \(I_\ell\) and the total mass of \(M\) of \(R\) as follows: \[k_{\ell}=\sqrt{\frac{I_{\ell}}{M}}\]
In practice, the following theorem is used to calculate the moment of inertia \(I_{\ell}\) with respect to a line \(\ell\).
Steiner's theorem The moment of inertia \(I_{\ell}\) with respect to a line \(\ell\) for a solid can be calculated as \[I_{\ell} = I_0+ Md^2\] where \(M\) is the mass of the solid, \(I_0\) is the moment of inertia with respect to the line \(\ell_0\) through the centroid and parallel to \(\ell\), and \(d\) is the distance between the two lines \(\ell\) and \(\ell_0\).
Similar formulas hold for a two-dimensional bounded region by simply omitting \(z\). In Steiner's theorem, mass must be replaced by area.
If for a two-dimensional region \(R\) the continuous density \(\rho(x,y)\) is given in every point of \(R\), then for an area element with very small dimensions \(\dd (x,y)\) its mass is approximately equal to \(\rho(x,y)\,\dd(x, y)\). \[M=\iint_R \rho(x,y)\,\dd(x, y)\] therefore is true for the total mass \(M\) of \(R\)
Static moments in two dimensions The 2D static moments \(M_{y}\) and \(M_{x}\) of \(R\), respectively, relative to the \(x\) and \(y\) axes, also called first moments of area, are defined as follows:
\[\begin{aligned} M_{y} &=\iint_R x\,\rho(x,y)\,\dd(x,y)\\[0.25cm]
M_{x} &=\iiint_R y\,\rho(x,y)\,\dd(x,y)\end{aligned}\]
The 2D centre of mass, more commonly known as centre of gravity, \((\bar{x},\bar{y})\) of \(R\) is given by: \[\bar{x}=\frac{M_{y}}{M},\quad \bar{y}=\frac{M_{x}}{M}\]
Determine the centroid of the region \(R\) sketched below in the plane and emclosed by the curve \(y=x^2\) and the line \(y=x\), assuming that the density is a constant \(\rho\). The centre of gravity is indicated in the figure below with a red dot.
\(\cv{\bar{x} \\ \bar{y}} ={}\) \(\cv{{{1}\over{2}}\\ {{2}\over{5}}}\)
First we calculate the total mass \(M\): \[\begin{aligned} M &= \iint_R \rho\,\dd(x,y)\\[0.25cm] &= \rho\, \int_{x=0}^{x=1}\left(\int_{y=x^2}^{y=x}\dd y\right)\dd x\\[0.25cm] &= \rho \int_{x=0}^{x=1}\biggl[y\biggr]_{y=x^2}^{y=x}\\[0.25cm] &=\rho \int_{x=0}^{x=1} \left(x-x^2\right)\,\dd x\\[0.25cm] &= \rho\cdot \biggl[{{1}\over{2}}x^2-\frac{1}{3}x^3\biggr]_{x=0}^{x=1}\\[0.25cm] &= {{1}\over{6}}\rho \end{aligned}\] Next we determine in a similar way the static moments \(M_{y}\) and \(M_{x}\): \[\begin{aligned} M_{y} &= \iint_R x\,\rho\,\dd(x,y)\\[0.25cm] &= \rho\int_{x=0}^{x=1}\left(\int_{y=x^2}^{y=x}x\, \dd y\right)\dd x\\[0.25cm] &= \rho \int_{x=0}^{x=1}\biggl[x\,y\biggr]_{y=x^2}^{y=x}\;\dd x\\[0.25cm] &=\rho \int_{x=0}^{x=1} \left(x^2-x^3\right)\,\dd x\\[0.25cm] &= \rho\cdot \biggl[{{1}\over{3}}x^3-\frac{1}{4}x^4\biggr]_{x=0}^{x=1}\\[0.25cm] &= {{1}\over{12}}\,\rho\\[1.0cm] M_{x} &= \iint_R y\,\rho\,\dd(x,y)\\[0.25cm] &= \rho\int_{x=0}^{x=1}\left(\int_{y=x^2}^{y=x}y\, \dd y\right)\dd x\\[0.25cm] &= \rho \int_{x=0}^{x=1}\biggl[\frac{1}{2}y^2\biggr]_{y=x^2}^{y=x}\\[0.25cm] &=\rho \int_{x=0}^{x=1} \left({{1}\over{2}} x^2-\frac{1}{2}x^4\right)\,\dd x\\[0.25cm] &= \rho\cdot \biggl[{{1}\over{6}}x^3-\frac{1}{10}x^5\biggr]_{x=0}^{x=1}\\[0.25cm] &= {{1}\over{15}}\,\rho \end{aligned}\] The coordinates of the centre of gravity can now be calculated as quotients: \[\begin{aligned} \bar{x}&=\frac{M_{y}}{M}\\[0.25cm] &= \frac{{{1}\over{12}}\,\rho}{{{1}\over{6}}\,\rho}\\[0.25cm] &= {{1}\over{2}}\\[0.5cm] \bar{y}&=\frac{M_{x}}{M}\\[0.25cm] &= \frac{{{1}\over{15}}\,\rho}{{{1}\over{6}}\,\rho}\\[0.25cm] &= {{2}\over{5}}\end{aligned}\]
Moments of inertia in two dimensions The 2D moment of inertia \(M_{\ell}\) of \(R\) relative to. the line \(\ell\), also called area moment of inertia or second moment of area, is defined as follows: \[I_{\ell} = \iiint \delta(x,y)^2\,\rho(x,y)\,\dd(x,y,z)\] where \[\delta(x,y)^2=\text{distance of the point }(x,y)\text{ to the line }\ell\text.\] Two special cases are the moments of inertia relative to the coordinate axes: \[\begin{aligned} I_x &=\iint_R y^2\,\rho(x,y)\,\dd(x,y)\\[0.25cm]
I_{y} &=\iint_R x^2\,\rho(x,y,z)\,\dd(x,y)\end{aligned}\] The moment of inertia \(I_O\) about the origin, also called the polar moment of inertia, is defined as \[\begin{aligned}I_O&=I_x+I_y\\[0.25cm] &=\iint \left((x^2+y^2\right)\,\rho(x,y)\,\dd(x,y)\end{aligned}\]
