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Calculating Centers of Mass and Moments of Inertia

Use double integrals to locate the center of mass of a two-dimensional object.

Center of Mass in Two Dimensions

The center of mass is also known as the center of gravity if the object is in a uniform gravitational field. If the object has uniform density, the center of mass is the geometric center of the object, which is called the centroid. shows a point \(P\) as the center of mass of a lamina. The lamina is perfectly balanced about its center of mass.

To find the coordinates of the center of mass \(P(\overset{\text{-}}{x},\overset{\text{-}}{y})\) of a lamina, we need to find the moment \({M}_{x}\) of the lamina about the \(x\text{-axis}\) and the moment \({M}_{y}\) about the \(y\text{-axis}\text{.}\) We also need to find the mass \(m\) of the lamina. Then

\[\overset{\text{-}}{x}=\frac{{M}_{y}}{m}\ \text{and}\ \overset{\text{-}}{y}=\frac{{M}_{x}}{m}.\]

Refer to Moments and Centers of Mass for the definitions and the methods of single integration to find the center of mass of a one-dimensional object (for example, a thin rod). We are going to use a similar idea here except that the object is a two-dimensional lamina and we use a double integral.

If we allow a constant density function, then \(\overset{\text{-}}{x}=\frac{{M}_{y}}{m}\ \text{and}\ \overset{\text{-}}{y}=\frac{{M}_{x}}{m}\) give the centroid of the lamina.

Suppose that the lamina occupies a region \(R\) in the \(xy\text{-plane},\) and let \(\rho (x,y)\) be its density (in units of mass per unit area) at any point \((x,y).\) Hence, \(\rho (x,y)=\underset{\Delta A\to 0}{\text{lim}}\frac{\Delta m}{\Delta A},\) where \(\Delta m\) and \(\Delta A\) are the mass and area of a small rectangle containing the point \((x,y)\) and the limit is taken as the dimensions of the rectangle go to \(0\) (see the following figure).

Just as before, we divide the region \(R\) into tiny rectangles \({R}_{ij}\) with area \(\Delta A\) and choose \(({x}_{ij}^{*},{y}_{ij}^{*})\) as sample points. Then the mass \({m}_{ij}\) of each \({R}_{ij}\) is equal to \(\rho ({x}_{ij}^{*},{y}_{ij}^{*})\Delta A\) (). Let \(k\) and \(l\) be the number of subintervals in \(x\) and \(y,\) respectively. Also, note that the shape might not always be rectangular but the limit works anyway, as seen in previous sections.

Hence, the mass of the lamina is

\[m=\underset{k,l\to \infty }{\text{lim}}\sum _{i=1}^{k}\sum _{j=1}^{l}{m}_{ij}=\underset{k,l\to \infty }{\text{lim}}\sum _{i=1}^{k}\sum _{j=1}^{l}\rho ({x}_{ij}^{*},{y}_{ij}^{*})\Delta A=\underset{R}{∬}\rho (x,y)dA.\]

Condensed: the full section is in OpenStax Calculus Volume 3.

Moments of Inertia

For a clear understanding of how to calculate moments of inertia using double integrals, we need to go back to the general definition of moments and centers of mass in Section 6.6 of Volume 1. 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 that the moment of inertia of the subrectangle \({R}_{ij}\) about the \(x\text{-axis}\) is \({({y}_{ij}^{*})}^{2}\rho ({x}_{ij}^{*},{y}_{ij}^{*})\Delta A.\) Similarly, the moment of inertia of the subrectangle \({R}_{ij}\) about the \(y\text{-axis}\) is \({({x}_{ij}^{*})}^{2}\rho ({x}_{ij}^{*},{y}_{ij}^{*})\Delta A.\) 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

\[{I}_{x}=\underset{k,l\to \infty }{\text{lim}}{\sum _{i=1}^{k}\sum _{j=1}^{l}({y}_{ij}^{*})}^{2}{m}_{ij}=\underset{k,l\to \infty }{\text{lim}}{\sum _{i=1}^{k}\sum _{j=1}^{l}({y}_{ij}^{*})}^{2}\rho ({x}_{ij}^{*},{y}_{ij}^{*})\Delta A=\underset{R}{∬}{y}^{2}\rho (x,y)dA.\]

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

\[{I}_{y}=\underset{k,l\to \infty }{\text{lim}}{\sum _{i=1}^{k}\sum _{j=1}^{l}({x}_{ij}^{*})}^{2}{m}_{ij}=\underset{k,l\to \infty }{\text{lim}}{\sum _{i=1}^{k}\sum _{j=1}^{l}({x}_{ij}^{*})}^{2}\rho ({x}_{ij}^{*},{y}_{ij}^{*})\Delta A=\underset{R}{∬}{x}^{2}\rho (x,y)dA.\]

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

\[{I}_{0}={I}_{x}+{I}_{y}=\underset{R}{∬}({x}^{2}+{y}^{2})\rho (x,y)dA.\]

All these expressions can be written in polar coordinates by substituting \(x=r\ \text{cos}\ \theta ,\) \(y=r\ \text{sin}\ \theta ,\) and \(dA=r\ dr\ d\theta .\) For example, \({I}_{0}=\underset{R}{∬}{r}^{2}\rho (r\ \text{cos}\ \theta ,r\ \text{sin}\ \theta )dA.\)

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

Condensed: the full section is in OpenStax Calculus Volume 3.

Center of Mass and Moments of Inertia in Three Dimensions

All the expressions of double integrals discussed so far can be modified to become triple integrals.

Example

Try it.

Suppose that \(Q\) is a solid region bounded by \(x+2y+3z=6\) and the coordinate planes and has density \(\rho (x,y,z)={x}^{2}yz.\) Find the total mass.

Solution

The region \(Q\) is a tetrahedron () meeting the axes at the points \((6,0,0),(0,3,0),\) and \((0,0,2).\) To find the limits of integration, let \(z=0\) in the slanted plane \(z=\frac{1}{3}(6-x-2y).\) Then for \(x\) and \(y\) find the projection of \(Q\) onto the \(xy\text{-plane,}\) which is bounded by the axes and the line \(x+2y=6.\) Hence the mass is

\[m=\underset{Q}{∭}\rho (x,y,z)dV=\int _{x=0}^{x=6}\ \int _{y=0}^{y=1\text{/}2(6-x)}\ \int _{z=0}^{z=1\text{/}3(6-x-2y)}{x}^{2}yz\ dz\ dy\ dx=\frac{108}{35}\approx 3.086.\]

We conclude this section with an example of finding moments of inertia \({I}_{x},{I}_{y},\) and \({I}_{z}.\)

Condensed: the full section is in OpenStax Calculus Volume 3.

Key Concepts

Finding the mass, center of mass, moments, and moments of inertia in double integrals:

  • For a lamina \(R\) with a density function \(\rho (x,y)\) at any point \((x,y)\) in the plane, the mass is \(m=\underset{R}{∬}\rho (x,y)dA.\)
  • The moments about the \(x\text{-axis}\) and \(y\text{-axis}\) are
    \[{M}_{x}=\underset{R}{∬}y\rho (x,y)dA\ \text{and}\ {M}_{y}=\underset{R}{∬}x\rho (x,y)dA.\]
  • The center of mass is given by \(\overset{\text{-}}{x}=\frac{{M}_{y}}{m},\overset{\text{-}}{y}=\frac{{M}_{x}}{m}.\)
  • The center of mass becomes the centroid of the plane when the density is constant.
  • The moments of inertia about the \(x-\text{axis,}\) \(y-\text{axis,}\) and the origin are
    \[{I}_{x}=\underset{R}{∬}{y}^{2}\rho (x,y)dA,\ \ {I}_{y}=\underset{R}{∬}{x}^{2}\rho (x,y)dA,\ \text{and}\ {I}_{0}={I}_{x}+{I}_{y}=\underset{R}{∬}({x}^{2}+{y}^{2})\rho (x,y)dA.\]

Finding the mass, center of mass, moments, and moments of inertia in triple integrals:

  • For a solid object \(Q\) with a density function \(\rho (x,y,z)\) at any point \((x,y,z)\) in space, the mass is \(m=\underset{Q}{∭}\rho (x,y,z)dV.\)
  • The moments about the \(xy\text{-plane,}\) the \(xz\text{-plane,}\) and the \(yz\text{-plane}\) are
    \[{M}_{xy}=\underset{Q}{∭}z\rho (x,y,z)dV,\ {M}_{xz}=\underset{Q}{∭}y\rho (x,y,z)dV,\ {M}_{yz}=\underset{Q}{∭}x\rho (x,y,z)dV.\]
  • The center of mass is given by \(\overset{\text{-}}{x}=\frac{{M}_{yz}}{m},\overset{\text{-}}{y}=\frac{{M}_{xz}}{m},\overset{\text{-}}{z}=\frac{{M}_{xy}}{m}.\)
  • The center of mass becomes the centroid of the solid when the density is constant.
  • The moments of inertia about the \(yz\text{-plane,}\) the \(xz\text{-plane,}\) and the \(xy\text{-plane}\) are
    \[\begin{array}{l} \\ {I}_{x}=\underset{Q}{∭}({y}^{2}+{z}^{2})\rho (x,y,z)dV,{I}_{y}=\underset{Q}{∭}({x}^{2}+{z}^{2})\rho (x,y,z)dV, \\ {I}_{z}=\underset{Q}{∭}({x}^{2}+{y}^{2})\rho (x,y,z)dV.\end{array}\]

Key Equations

Mass of a lamina\(m=\underset{k,l\to \infty }{\text{lim}}\sum _{i=1}^{k}\sum _{j=1}^{l}{m}_{ij}=\underset{k,l\to \infty }{\text{lim}}\sum _{i=1}^{k}\sum _{j=1}^{l}\rho ({x}_{ij}^{*},{y}_{ij}^{*})\Delta A=\underset{R}{∬}\rho (x,y)dA\)
Moment about the x-axis\({M}_{x}=\underset{k,l\to \infty }{\text{lim}}\sum _{i=1}^{k}\sum _{j=1}^{l}({y}_{ij}^{*}){m}_{ij}=\underset{k,l\to \infty }{\text{lim}}\sum _{i=1}^{k}\sum _{j=1}^{l}({y}_{ij}^{*})\rho ({x}_{ij}^{*},{y}_{ij}^{*})\Delta A=\underset{R}{∬}y\rho (x,y)dA\)
Moment about the y-axis\({M}_{y}=\underset{k,l\to \infty }{\text{lim}}\sum _{i=1}^{k}\sum _{j=1}^{l}({x}_{ij}^{*}){m}_{ij}=\underset{k,l\to \infty }{\text{lim}}\sum _{i=1}^{k}\sum _{j=1}^{l}({x}_{ij}^{*})\rho ({x}_{ij}^{*},{y}_{ij}^{*})\Delta A=\underset{R}{∬}x\rho (x,y)dA\)
Center of mass of a lamina\(\overset{\text{-}}{x}=\frac{{M}_{y}}{m}=\frac{\underset{R}{∬}x\rho (x,y)dA}{\underset{R}{∬}\rho (x,y)dA}\) and \(\overset{\text{-}}{y}=\frac{{M}_{x}}{m}=\frac{\underset{R}{∬}y\rho (x,y)dA}{\underset{R}{∬}\rho (x,y)dA}\)

Calculating Centers of Mass and Moments of Inertia

In the following exercises, the region \(R\) occupied by a lamina is shown in a graph. Find the mass of \(R\) with the density function \(\rho .\)

In the following exercises, consider a lamina occupying the region \(R\) and having the density function \(\rho\) given in the preceding group of exercises. Use a computer algebra system (CAS) to answer the following questions.

  1. Find the moments \({M}_{x}\) and \({M}_{y}\) about the \(x\text{-axis}\) and \(y\text{-axis,}\) respectively.
  2. Calculate and plot the center of mass of the lamina.
  3. [T] Use a CAS to locate the center of mass on the graph of \(R.\)

In the following exercises, consider a lamina occupying the region \(R\) and having the density function \(\rho\) given in the first two groups of Exercises.

  1. Find the moments of inertia \({I}_{x},{I}_{y},\) and \({I}_{0}\) about the \(x\text{-axis},\) \(y\text{-axis},\) and origin, respectively.
  2. Find the radii of gyration with respect to the \(x\text{-axis},\) \(y\text{-axis},\) and origin, respectively.

이제 너 계산기는 이것을 해결하지 않지만, 그 조각은 계산 가능합니다. 아래의 하나를 시도하거나 자신의 것을 입력하십시오.

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  1. Consider a triangular lamina \(R\) with vertices \((0,0),(0,3),\) \((3,0)\) and with density \(\rho (x,y)=xy{\ \text{kg/m}}^{2}.\) Find the total mass.

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    A sketch of the region \(R\) is always helpful, as shown in the following figure.

    Using the expression developed for mass, we see that

    \[\begin{array}{ll}m & =\underset{R}{∬}dm=\underset{R}{∬}\rho (x,y)dA=\int _{x=0}^{x=3}\ \int _{y=0}^{y=3-x}xy\ dy\ dx=\int _{x=0}^{x=3}[{x\frac{{y}^{2}}{2}|}_{y=0}^{y=3-x}]dx \\ & =\int _{x=0}^{x=3}\frac{1}{2}x{(3-x)}^{2}dx={[\frac{9{x}^{2}}{4}-{x}^{3}+\frac{{x}^{4}}{8}]|}_{x=0}^{x=3} \\ & =\frac{27}{8}.\end{array}\]

    The computation is straightforward, giving the answer \(m=\frac{27}{8}\ \text{kg}\text{.}\)

  2. Consider the same region \(R\) as in the previous example, and use the density function \(\rho (x,y)=\sqrt{xy}.\) Find the total mass. Hint: Use trigonometric substitution \(\sqrt{x}=\sqrt{3}\sin \theta\) and then use the power reducing formulas for trigonometric functions.

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    \(\frac{9\pi }{8}\ \text{kg}\)

  3. Consider the same triangular lamina \(R\) with vertices \((0,0),(0,3),\ (3,0)\) and with density \(\rho (x,y)=xy.\) Find the moments \({M}_{x}\) and \({M}_{y}.\)

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    Use double integrals for each moment and compute their values:

    \[{M}_{x}=\underset{R}{∬}y\rho (x,y)dA=\int _{x=0}^{x=3}\ \int _{y=0}^{y=3-x}x{y}^{2}\ dy\ dx=\frac{81}{20},\]\[{M}_{y}=\underset{R}{∬}x\rho (x,y)dA=\int _{x=0}^{x=3}\ \int _{y=0}^{y=3-x}{x}^{2}\ yd\ y\ dx=\frac{81}{20}.\]

    The computation is quite straightforward.

  4. Consider the same lamina \(R\) as above, and use the density function \(\rho (x,y)=\sqrt{xy}.\) Find the moments \({M}_{x}\) and \({M}_{y}.\)

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    \({M}_{x}=\frac{81\pi }{64}\) and \({M}_{y}=\frac{81\pi }{64}\)

  5. Again consider the same triangular region \(R\) with vertices \((0,0),(0,3),\) \((3,0)\) and with density function \(\rho (x,y)=xy.\) Find the center of mass.

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    Using the formulas we developed, we have

    \[\overset{\text{-}}{x}=\frac{{M}_{y}}{m}=\frac{\underset{R}{∬}x\rho (x,y)dA}{\underset{R}{∬}\rho (x,y)dA}=\frac{81\text{/}20}{27\text{/}8}=\frac{6}{5},\]\[\overset{\text{-}}{y}=\frac{{M}_{x}}{m}=\frac{\underset{R}{∬}y\rho (x,y)dA}{\underset{R}{∬}\rho (x,y)dA}=\frac{81\text{/}20}{27\text{/}8}=\frac{6}{5}.\]

    Therefore, the center of mass is the point \((\frac{6}{5},\frac{6}{5}).\)

  6. Again use the same region \(R\) as above and the density function \(\rho (x,y)=\sqrt{xy}.\) Find the center of mass.

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    \(\overset{\text{-}}{x}=\frac{{M}_{y}}{m}=\frac{81\pi \text{/}64}{9\pi \text{/}8}=\frac{9}{8}\) and \(\overset{\text{-}}{y}=\frac{{M}_{x}}{m}=\frac{81\pi \text{/}64}{9\pi \text{/}8}=\frac{9}{8}.\)

  7. Find the mass, moments, and the center of mass of the lamina of density \(\rho (x,y)=x+y\) occupying the region \(R\) under the curve \(y={x}^{2}\) in the interval \(0\le x\le 2\) (see the following figure).

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    First we compute the mass \(m.\) We need to describe the region between the graph of \(y={x}^{2}\) and the vertical lines \(x=0\) and \(x=2\text{:}\)

    \[\begin{array}{ll}m & =\underset{R}{∬}dm=\underset{R}{∬}\rho (x,y)dA=\int _{x=0}^{x=2}\ \int _{y=0}^{y={x}^{2}}(x+y)dy\ dx=\int _{x=0}^{x=2}[{xy+\frac{{y}^{2}}{2}|}_{y=0}^{y={x}^{2}}]dx \\ & =\int _{x=0}^{x=2}[{x}^{3}+\frac{{x}^{4}}{2}]dx={[\frac{{x}^{4}}{4}+\frac{{x}^{5}}{10}]|}_{x=0}^{x=2}=\frac{36}{5}.\end{array}\]

    Now compute the moments \({M}_{x}\) and \({M}_{y}\text{:}\)

    \[{M}_{x}=\underset{R}{∬}y\rho (x,y)dA=\int _{x=0}^{x=2}\ \int _{y=0}^{y={x}^{2}}y(x+y)dy\ dx=\frac{80}{7},\]\[{M}_{y}=\underset{R}{∬}x\rho (x,y)dA=\int _{x=0}^{x=2}\ \int _{y=0}^{y={x}^{2}}x(x+y)dy\ dx=\frac{176}{15}.\]

    Finally, evaluate the center of mass,

    \[\begin{array}{l} \\ \\ \\ \overset{\text{-}}{x}=\frac{{M}_{y}}{m}=\frac{\underset{R}{∬}x\rho (x,y)dA}{\underset{R}{∬}\rho (x,y)dA}=\frac{176\text{/}15}{36\text{/}5}=\frac{44}{27}, \\ \overset{\text{-}}{y}=\frac{{M}_{x}}{m}=\frac{\underset{R}{∬}y\rho (x,y)dA}{\underset{R}{∬}\rho (x,y)dA}=\frac{80\text{/}7}{36\text{/}5}=\frac{100}{63}.\end{array}\]

    Hence the center of mass is \((\overset{\text{-}}{x},\overset{\text{-}}{y})=(\frac{44}{27},\frac{100}{63}).\)

  8. Calculate the mass, moments, and the center of mass of the region between the curves \(y=x\) and \(y={x}^{2}\) with the density function \(\rho (x,y)=x\) in the interval \(0\le x\le 1.\)

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    \(\overset{\text{-}}{x}=\frac{{M}_{y}}{m}=\frac{1\text{/}20}{1\text{/}12}=\frac{3}{5}\) and \(\overset{\text{-}}{y}=\frac{{M}_{x}}{m}=\frac{1\text{/}24}{1\text{/}12}=\frac{1}{2}\)

  9. Find the centroid of the region under the curve \(y={e}^{x}\) over the interval \(1\le x\le 3\) (see the following figure).

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    To compute the centroid, we assume that the density function is constant and hence it cancels out:

    \[\begin{array}{l} \\ \\ \\ \\ {x}_{c}=\frac{{M}_{y}}{m}=\frac{\underset{R}{∬}x\ dA}{\underset{R}{∬}dA}\ \text{and}\ {y}_{c}=\frac{{M}_{x}}{m}=\frac{\underset{R}{∬}y\ dA}{\underset{R}{∬}dA}, \\ {x}_{c}=\frac{{M}_{y}}{m}=\frac{\underset{R}{∬}x\ dA}{\underset{R}{∬}dA}=\frac{\int _{x=1}^{x=3}\ \int _{y=0}^{y={e}^{x}}x\ dy\ dx}{\int _{x=1}^{x=3}\ \int _{y=0}^{y={e}^{x}}dy\ dx}=\frac{\int _{x=1}^{x=3}x{e}^{x}dx}{\int _{x=1}^{x=3}{e}^{x}dx}=\frac{2{e}^{3}}{{e}^{3}-e}=\frac{2{e}^{2}}{{e}^{2}-1}, \\ {y}_{c}=\frac{{M}_{x}}{m}=\frac{\underset{R}{∬}y\ dA}{\underset{R}{∬}dA}=\frac{\int _{x=1}^{x=3}\ \int _{y=0}^{y={e}^{x}}y\ dy\ dx}{\int _{x=1}^{x=3}\ \int _{y=0}^{y={e}^{x}}dy\ dx}=\frac{\int _{x=1}^{x=3}\frac{{e}^{2x}}{2}dx}{\int _{x=1}^{x=3}{e}^{x}dx}=\frac{\frac{1}{4}{e}^{2}({e}^{4}-1)}{e({e}^{2}-1)}=\frac{1}{4}e({e}^{2}+1).\end{array}\]

    Thus the centroid of the region is

    \[({x}_{c},{y}_{c})=(\frac{2{e}^{2}}{{e}^{2}-1},\frac{1}{4}e({e}^{2}+1)).\]

  10. Calculate the centroid of the region between the curves \(y=x\) and \(y=\sqrt{x}\) with uniform density in the interval \(0\le x\le 1.\)

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    \({x}_{c}=\frac{{M}_{y}}{m}=\frac{1\text{/}15}{1\text{/}6}=\frac{2}{5}\ \text{and}\ {y}_{c}=\frac{{M}_{x}}{m}=\frac{1\text{/}12}{1\text{/}6}=\frac{1}{2}\)

  11. Use the triangular region \(R\) with vertices \((0,0),(2,2),\) and \((2,0)\) and with density \(\rho (x,y)=xy\) as in previous examples. Find the moments of inertia.

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    Using the expressions established above for the moments of inertia, we have

    \[\begin{array}{lll}{I}_{x} & = & \underset{R}{∬}{y}^{2}\rho (x,y)dA=\int _{x=0}^{x=2}\ \int _{y=0}^{y=x}x{y}^{3}dy\ dx=\frac{8}{3}, \\ {I}_{y} & = & \underset{R}{∬}{x}^{2}\rho (x,y)dA=\int _{x=0}^{x=2}\ \int _{y=0}^{y=x}{x}^{3}y\ dy\ dx=\frac{16}{3}, \\ {I}_{0} & = & \underset{R}{∬}({x}^{2}+{y}^{2})\rho (x,y)dA=\int _{0}^{2}\ \int _{0}^{x}({x}^{2}+{y}^{2})xy\ dy\ dx \\ & = & {I}_{x}+{I}_{y}=8.\end{array}\]

  12. Again use the same region \(R\) as above and the density function \(\rho (x,y)=\sqrt{xy}.\) Find the moments of inertia.

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    \({I}_{x}=\int _{x=0}^{x=2}\ \int _{y=0}^{y=x}{y}^{2}\sqrt{xy}\ dy\ dx=\frac{64}{35}\) and \({I}_{y}=\int _{x=0}^{x=2}\ \int _{y=0}^{y=x}{x}^{2}\sqrt{xy}\ dy\ dx=\frac{64}{35}.\) Also, \({I}_{0}=\int _{x=0}^{x=2}\ \int _{y=0}^{y=x}({x}^{2}+{y}^{2})\sqrt{xy}\ dy\ dx=\frac{128}{35}.\)

  13. Consider the same triangular lamina \(R\) with vertices \((0,0),(2,2),\) and \((2,0)\) and with density \(\rho (x,y)=xy\) as in previous examples. Find the radii of gyration with respect to the \(x\text{-axis,}\) the \(y\text{-axis,}\) and the origin.

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    If we compute the mass of this region we find that \(m=2.\) We found the moments of inertia of this lamina in . From these data, the radii of gyration with respect to the \(x\text{-axis,}\) \(y\text{-axis,}\) and the origin are, respectively,

    \[\begin{array}{lll}{R}_{x} & = & \sqrt{\frac{{I}_{x}}{m}}=\sqrt{\frac{8\text{/}3}{2}}=\sqrt{\frac{8}{6}}=\frac{2\sqrt{3}}{3}, \\ {R}_{y} & = & \sqrt{\frac{{I}_{y}}{m}}=\sqrt{\frac{16\text{/}3}{2}}=\sqrt{\frac{8}{3}}=\frac{2\sqrt{6}}{3}, \\ {R}_{0} & = & \sqrt{\frac{{I}_{0}}{m}}=\sqrt{\frac{8}{2}}=\sqrt{4}=2.\end{array}\]

  14. Use the same region \(R\) from and the density function \(\rho (x,y)=\sqrt{xy}.\) Find the radii of gyration with respect to the \(x\text{-axis,}\) the \(y\text{-axis,}\) and the origin.

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    \({R}_{x}=\frac{6\sqrt{35}}{35},\) \({R}_{y}=\frac{6\sqrt{35}}{35},\) and \({R}_{0}=\frac{6\sqrt{70}}{35}.\)

  15. Suppose that \(Q\) is a solid region bounded by \(x+2y+3z=6\) and the coordinate planes and has density \(\rho (x,y,z)={x}^{2}yz.\) Find the total mass.

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    The region \(Q\) is a tetrahedron () meeting the axes at the points \((6,0,0),(0,3,0),\) and \((0,0,2).\) To find the limits of integration, let \(z=0\) in the slanted plane \(z=\frac{1}{3}(6-x-2y).\) Then for \(x\) and \(y\) find the projection of \(Q\) onto the \(xy\text{-plane,}\) which is bounded by the axes and the line \(x+2y=6.\) Hence the mass is

    \[m=\underset{Q}{∭}\rho (x,y,z)dV=\int _{x=0}^{x=6}\ \int _{y=0}^{y=1\text{/}2(6-x)}\ \int _{z=0}^{z=1\text{/}3(6-x-2y)}{x}^{2}yz\ dz\ dy\ dx=\frac{108}{35}\approx 3.086.\]

  16. Consider the same region \(Q\) (), and use the density function \(\rho (x,y,z)=x{y}^{2}z.\) Find the mass.

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    \(\frac{54}{35}=1.543\)

  17. Suppose \(Q\) is a solid region bounded by the plane \(x+2y+3z=6\) and the coordinate planes with density \(\rho (x,y,z)={x}^{2}yz\) (see ). Find the center of mass using decimal approximation. Use the mass found in

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    We have used this tetrahedron before and know the limits of integration, so we can proceed to the computations right away. First, we need to find the moments about the \(xy\text{-plane,}\) the \(xz\text{-plane,}\) and the \(yz\text{-plane:}\)

    \[\begin{array}{l} \\ \\ \\ {M}_{xy}=\underset{Q}{∭}z\rho (x,y,z)dV=\int _{x=0}^{x=6}\ \int _{y=0}^{y=1\text{/}2(6-x)}\ \int _{z=0}^{z=1\text{/}3(6-x-2y)}{x}^{2}y{z}^{2}dz\ dy\ dx=\frac{54}{35}\approx 1.543, \\ {M}_{xz}=\underset{Q}{∭}y\rho (x,y,z)dV=\int _{x=0}^{x=6}\ \int _{y=0}^{y=1\text{/}2(6-x)}\ \int _{z=0}^{z=1\text{/}3(6-x-2y)}{x}^{2}{y}^{2}z\ dz\ dy\ dx=\frac{81}{35}\approx 2.314, \\ {M}_{yz}=\underset{Q}{∭}x\rho (x,y,z)dV=\int _{x=0}^{x=6}\ \int _{y=0}^{y=1\text{/}2(6-x)}\ \int _{z=0}^{z=1\text{/}3(6-x-2y)}{x}^{3}yz\ dz\ dy\ dx=\frac{243}{35}\approx 6.943.\end{array}\]

    Hence the center of mass is

    \[\begin{array}{l} \\ \\ \\ \overset{\text{-}}{x}=\frac{{M}_{yz}}{m},\overset{\text{-}}{y}=\frac{{M}_{xz}}{m},\overset{\text{-}}{z}=\frac{{M}_{xy}}{m}, \\ \overset{\text{-}}{x}=\frac{{M}_{yz}}{m}=\frac{243\text{/}35}{108\text{/}35}=\frac{243}{108}=2.25, \\ \overset{\text{-}}{y}=\frac{{M}_{xz}}{m}=\frac{81\text{/}35}{108\text{/}35}=\frac{81}{108}=0.75, \\ \overset{\text{-}}{z}=\frac{{M}_{xy}}{m}=\frac{54\text{/}35}{108\text{/}35}=\frac{54}{108}=0.5.\end{array}\]

    The center of mass for the tetrahedron \(Q\) is the point \((2.25,0.75,0.5).\)

  18. Consider the same region \(Q\) () and use the density function \(\rho (x,y,z)=x{y}^{2}z.\) Find the center of mass.

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    \((\frac{3}{2},\frac{9}{8},\frac{1}{2})\)

  19. Suppose that \(Q\) is a solid region and is bounded by \(x+2y+3z=6\) and the coordinate planes with density \(\rho (x,y,z)={x}^{2}yz\) (see ). Find the moments of inertia of the tetrahedron \(Q\) about the \(yz\text{-plane,}\) the \(xz\text{-plane,}\) and the \(xy\text{-plane}\text{.}\)

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    Once again, we can almost immediately write the limits of integration and hence we can quickly proceed to evaluating the moments of inertia. Using the formula stated before, the moments of inertia of the tetrahedron \(Q\) about the \(xy\text{-plane,}\) the \(xz\text{-plane,}\) and the \(yz\text{-plane}\) are

    \[\begin{array}{l} \\ {I}_{x}=\underset{Q}{∭}({y}^{2}+{z}^{2})\rho (x,y,z)dV, \\ {I}_{y}=\underset{Q}{∭}({x}^{2}+{z}^{2})\rho (x,y,z)dV,\end{array}\]

    and

    \[{I}_{z}=\underset{Q}{∭}({x}^{2}+{y}^{2})\rho (x,y,z)dV\ \text{with}\ \rho (x,y,z)={x}^{2}yz.\]

    Proceeding with the computations, we have

    \[\begin{array}{l} \\ \\ \\ \\ \\ \\ {I}_{x}=\underset{Q}{∭}({y}^{2}+{z}^{2}){x}^{2}yz\ dV=\int _{x=0}^{x=6}\ \int _{y=0}^{y=\frac{1}{2}(6-x)}\ \int _{z=0}^{z=\frac{1}{3}(6-x-2y)}({y}^{2}+{z}^{2}){x}^{2}yz\ dz\ dy\ dx=\frac{117}{35}\approx 3.343, \\ {I}_{y}=\underset{Q}{∭}({x}^{2}+{z}^{2}){x}^{2}yz\ dV=\int _{x=0}^{x=6}\ \int _{y=0}^{y=\frac{1}{2}(6-x)}\ \int _{z=0}^{z=\frac{1}{3}(6-x-2y)}({x}^{2}+{z}^{2}){x}^{2}yz\ dz\ dy\ dx=\frac{684}{35}\approx 19.543, \\ {I}_{z}=\underset{Q}{∭}({x}^{2}+{y}^{2}){x}^{2}yz\ dV=\int _{x=0}^{x=6}\ \int _{y=0}^{y=\frac{1}{2}(6-x)}\ \int _{z=0}^{z=\frac{1}{3}(6-x-2y)}({x}^{2}+{y}^{2}){x}^{2}yz\ dz\ dy\ dx=\frac{729}{35}\approx 20.829.\end{array}\]

    Thus, the moments of inertia of the tetrahedron \(Q\) about the \(yz\text{-plane,}\) the \(xz\text{-plane,}\) and the \(xy\text{-plane}\) are \(117\text{/}35,684\text{/}35,\text{and}\ 729\text{/}35,\) respectively.

  20. Consider the same region \(Q\) (), and use the density function \(\rho (x,y,z)=x{y}^{2}z.\) Find the moments of inertia about the three coordinate planes.

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    The moments of inertia of the tetrahedron \(Q\) about the \(yz\text{-plane,}\) the \(xz\text{-plane,}\) and the \(xy\text{-plane}\) are \(99\text{/}35,36\text{/}7,\text{and}\ 243\text{/}35,\) respectively.

  21. \(R\) is the triangular region with vertices \((0,0),(0,3),\) and \((6,0);\rho (x,y)=xy.\)

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    \(\frac{27}{2}\)

  22. \(R\) is the triangular region with vertices \((0,0),(1,1),\) \((0,5);\rho (x,y)=x+y.\)

  23. \(R\) is the rectangular region with vertices \((0,0),(0,3),(6,3),\) and \((6,0);\) \(\rho (x,y)=\sqrt{xy}.\)

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    \(24\sqrt{2}\)

  24. \(R\) is the rectangular region with vertices \((0,1),(0,3),(3,3),\) and \((3,1);\) \(\rho (x,y)={x}^{2}y.\)

  25. \(R\) is the trapezoidal region determined by the lines \(y=-\frac{1}{4}x+\frac{5}{2},y=0,y=2,\) and \(x=0;\) \(\rho (x,y)=3xy.\)

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    \(76\)

  26. \(R\) is the trapezoidal region determined by the lines \(y=0,y=1,y=x,\) and \(y=\text{-}x+3;\rho (x,y)=2x+y.\)

  27. \(R\) is the disk of radius \(2\) centered at \((1,2);\) \(\rho (x,y)={x}^{2}+{y}^{2}-2x-4y+5.\)

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    \(8\pi\)

  28. \(R\) is the unit disk; \(\rho (x,y)=3{x}^{4}+6{x}^{2}{y}^{2}+3{y}^{4}.\)

  29. \(R\) is the region enclosed by the ellipse \({x}^{2}+4{y}^{2}=1;\rho (x,y)=1.\)

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    \(\frac{\pi }{2}\)

  30. \(R=\{(x,y)|9{x}^{2}+{y}^{2}\le 1,x\ge 0,y\ge 0\};\) \(\rho (x,y)=\sqrt{9{x}^{2}+{y}^{2}}.\)

  31. \(R\) is the region bounded by \(y=x,y=\text{-}x,y=x+2,y=\text{-}x+2;\) \(\rho (x,y)=1.\)

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    \(2\)

  32. \(R\) is the region bounded by \(y=\frac{1}{x},y=\frac{2}{x},y=1,\) and \(y=2;\rho (x,y)=4(x+y).\)

  33. [T] \(R\) is the triangular region with vertices \((0,0),(0,3),\) and \((6,0);\rho (x,y)=xy.\)

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    a. \({M}_{x}=\frac{81}{5},{M}_{y}=\frac{162}{5};\) b. \(\overset{\text{-}}{x}=\frac{12}{5},\overset{\text{-}}{y}=\frac{6}{5};\)
    c.

  34. [T] \(R\) is the triangular region with vertices \((0,0),(1,1),\text{and}\ (0,5);\rho (x,y)=x+y.\)

  35. [T] \(R\) is the rectangular region with vertices \((0,0),(0,3),(6,3),\text{and}\ (6,0);\) \(\rho (x,y)=\sqrt{xy}.\)

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    a. \({M}_{x}=\frac{216\sqrt{2}}{5},{M}_{y}=\frac{432\sqrt{2}}{5};\) b. \(\overset{\text{-}}{x}=\frac{18}{5},\overset{\text{-}}{y}=\frac{9}{5};\)
    c.

  36. [T] \(R\) is the rectangular region with vertices \((0,1),(0,3),(3,3),\text{and}\ (3,1);\) \(\rho (x,y)={x}^{2}y.\)

  37. [T] \(R\) is the trapezoidal region determined by the lines \(y=-\frac{1}{4}x+\frac{5}{2},y=0,\) \(y=2,\text{and}\ x=0;\) \(\rho (x,y)=3xy.\)

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    a. \({M}_{x}=\frac{368}{5},{M}_{y}=\frac{1552}{5};\) b. \(\overset{\text{-}}{x}=\frac{388}{95},\overset{\text{-}}{y}=\frac{92}{95};\)
    c.

  38. [T] \(R\) is the trapezoidal region determined by the lines \(y=0,y=1,y=x,\) and \(y=\text{-}x+3;\rho (x,y)=2x+y.\)

  39. [T] \(R\) is the disk of radius \(2\) centered at \((1,2);\) \(\rho (x,y)={x}^{2}+{y}^{2}-2x-4y+5.\)

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    a. \({M}_{x}=16\pi ,{M}_{y}=8\pi ;\) b. \(\overset{\text{-}}{x}=1,\overset{\text{-}}{y}=2;\)
    c.

  40. [T] \(R\) is the unit disk; \(\rho (x,y)=3{x}^{4}+6{x}^{2}{y}^{2}+3{y}^{4}.\)

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어떻게: Calculating Centers of Mass and Moments of Inertia

  1. Use double integrals to locate the center of mass of a two-dimensional object.
  2. Use double integrals to find the moment of inertia of a two-dimensional object.
  3. Use triple integrals to locate the center of mass of a three-dimensional object.
  4. For a lamina
  5. The moments about the
  6. The center of mass is given by
  7. The center of mass becomes the centroid of the plane when the density is constant.
  8. The moments of inertia about the

사람들이 묻는 질문

What is a partial derivative?

The ordinary derivative with respect to one variable while every other variable is frozen: the slope of the surface in one coordinate direction.

What does the gradient point at?

Uphill: the direction of steepest increase, with length equal to that steepest slope. It is perpendicular to the level curves.

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에 더 Multivariable Calculus

Functions of several variablesPartial derivatives and the gradientOptimisation in several variablesDouble and triple integralsVector fields, line integrals and the big theorems