Suppose a man and his son balance on a seesaw. If the man weights 200 lb and his son weighs 50 lb, where is the balance point on a 10 ft seesaw? Recall that the magnitude of the moment that they each produce around the balance point is defined as the product of the force they exert and their perpendicular distance from the balance point. In order to balance, these two moments must be equal:

\[\begin{align} W_{father} \cdot d_{father} &= W_{son} \cdot d_{son}\\ 200x &= 50 (10-x)\\ 200x &= 500 - 50x\\ 250x &= 500\\ x &= 2\ ft \end{align}\]Therefore, in order to balance, the balance point should be placed 2 ft from the father's end of the seesaw. This balance point is called the **center of mass**.

Take a look at the more general case, where \(\overline{x}\) is the position of the center of mass:

Equate the moments and solve for the location of \(\overline{x}\):

\[\begin{align} m_1 (\overline{x}-x_1) &= m_2 (x_2 - \overline{x})\\ m_1 \overline{x} + m_2 \overline{x} &= m_1 x_1 + m_2 x_2\\ \\ \overline{x} &= \dfrac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \end{align}\]Thus, in the one-dimensional case with discrete masses,

\[\ans{\overline{x} = \dfrac{\sum m_i x_i}{\sum m_i}}\]What if, instead of a bar with clear distinct masses resting on it, we have something like the situation below, where we're trying to balance an object, and the center of mass depends on its shape?

In this case, instead of using a sum to add up finitely many masses and distances, we'll use an integral (replace each sum with an integral, and replace the discrete masses with a density function \(\rho (x)\)):

\[\ans{\overline{x} = \dfrac{\int_a^b x \rho(x)\ dx}{\int_a^b \rho(x)\ dx} = \dfrac{1}{m} \int_a^b x \rho (x)\ dx}\] where \(m\) is the total mass (which can be found by integrating the density function), and \(\displaystyle\int_a^b x\rho(x)\ dx\) is called theConsider the following figure, where we're looking down on a thin sheet in two dimensions, and we assume that this sheet has varying density given by a density function \(\rho\).

Now, there will be two coordinates for the center of mass, or the *centroid*. These can be found by using similar integrals to what we saw before, but we'll use double integrals this time:

Consider the region \(R\) bounded by the y-axis and the curves \(y=e^{-x}-\dfrac{1}{2}\) and \(y=\dfrac{1}{2}-e^{-x}\). If the density of this sheet is \(\rho(x,y) = 1\), find its center of mass.

Start by calculating \(m\), since we'll need it for both formulas. Remember that the mass is the integral of the density function:

\[\begin{align} m &= \iint_R \rho(x,y)\ dA\\ &= \int_0^{\ln 2} \int_{1/2-e^{-x}}^{e^{-x-1/2}} \ dy\ dx\\ &= \int_0^{\ln 2} -2e^{-x}-x \bigg|_0^{\ln 2}\\ &= 1-\ln 2 \end{align}\]Next, apply the formulas to find the center of mass:

\[\begin{align} \overline{x} &= \dfrac{1}{1-\ln 2} \int_0^{\ln 2} \int_{1/2-e^{-x}}^{e^{-x-1/2}} x\ dy\ dx\\ &= \dfrac{1}{1-\ln 2} \int_0^{\ln 2} 2xe^{-x} - x\ dx\\ &= \dfrac{1}{1-\ln 2} \bigg[ -2xe^{-x}-\dfrac{1}{2}x^2-2e^{-x} \bigg]_0^{\ln 2}\\ &= \dfrac{1-\ln 2 - \dfrac{1}{2}(\ln 2)^2}{1-\ln 2} \approx 0.217\\ \overline{y} &= \dfrac{1}{1-\ln 2} \int_0^{\ln 2} \int_{1/2-e^{-x}}^{e^{-x-1/2}} y\ dy\ dx\\ &= \dfrac{1}{1-\ln 2} \int_0^{\ln 2} \dfrac{1}{2}\bigg[\left(e^{-x}-\dfrac{1}{2}\right)^2 - \left(\dfrac{1}{2}-e^{-x}\right)^2\bigg]\ dx\\ &= \dfrac{1}{2(1-\ln 2)} \int_0^{\ln 2} e^{-2x}-e^{-x}+\dfrac{1}{4}-\dfrac{1}{4}-e^{-2x}+e^{-x}\ dx = 0\\ \end{align}\]Therefore, the center of mass of this sheet is located at approximately

\[\ans{(\overline{x},\overline{y}) = (0.217,0)}\]It shouldn't be surprising that the y component of the center of mass is 0, because this sheet is symmetric about the x-axis.

Consider the region \(R=\{(x,y)\ :\ -1 \leq x \leq 1, 0 \leq y \leq 1\}\). If the density of this sheet is \(\rho(x,y) = 2-y\), find its center of mass.

Again, start by calculating \(m\):

\[\begin{align} m &= \iint_R \rho(x,y)\ dA\\ &= \int_{-1}^{1} \int_0^1 2-y \ dy\ dx\\ &= \int_{-1}^1 \dfrac{3}{2} \ dx = \dfrac{3}{2}x\bigg|_{-1}^1\\ &= 3 \end{align}\]Next, apply the formulas to find the center of mass:

\[\begin{align} \overline{x} &= \dfrac{1}{3} \int_{-1}^{1} \int_0^1 x(2-y) \ dy\ dx\\ &= \dfrac{1}{3} \int_{-1}^{1} \dfrac{3}{2}x\ dx\\ &= \dfrac{1}{3} \left(\dfrac{3}{4}x^2\bigg|_{-1}^1\right) = 0\\ \overline{y} &= \dfrac{1}{3} \int_{-1}^{1} \int_0^1 y(2-y) \ dy\ dx\\ &= \dfrac{1}{3} \int_{-1}^{1} y^2-\dfrac{1}{3}y^3\bigg|_0^1\ dx\\ &= \dfrac{1}{3} \int_{-1}^{1} \dfrac{2}{3}\ dx = \dfrac{4}{9}\\ \end{align}\]Therefore, the center of mass of this sheet is located at

\[\ans{(\overline{x},\overline{y}) = \left(0,\dfrac{4}{9}\right)}\]- Consider the region \(R=\{(x,y)\ :\ 0 \leq x \leq 4, 0 \leq y \leq 2\}\). If the density of this sheet is \(\rho(x,y) = 1+\dfrac{x}{2}\), find its center of mass.
- Find the center of mass of the triangular plate in the first quadrant bounded by \(x+y=4\) with \(\rho(x,y)=1+x+y\).

\((\overline{x},\overline{y}) = \left(\dfrac{7}{3},1\right)\)

\((\overline{x},\overline{y}) = \left(\dfrac{16}{11},\dfrac{16}{11}\right)\)

The center of mass in three dimensions is similar, but now we add a third component, and the integrals become triple integrals. If \[m=\iiint_D \rho(x,y,z)\ dV,\] then

\[\ans{\begin{align} \overline{x} &= \dfrac{1}{m} \iiint_D x \rho(x,y,z) \ dV\\ \\ \overline{y} &= \dfrac{1}{m} \iiint_D y \rho(x,y,z) \ dV\\ \\ \overline{z} &= \dfrac{1}{m} \iiint_D z \rho(x,y,z) \ dV \end{align}}\]Consider the cone \(D\) bounded by \(z=4-\sqrt{x^2+y^2}\) and \(z=0\). If the density of this cone is \(\rho(x,y,z) = 1\), find its center of mass.

As always, start by calculating \(m\). Let's try it in rectangular coordinates, but we'll get stuck pretty quickly:

\[\begin{align} m &= \iiint_D \rho(x,y,z)\ dV\\ &= \int_{-4}^{4} \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \int_0^{4-\sqrt{x^2+y^2}} 1\ dz\ dy\ dx\\ &= \int_{-4}^{4} \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} 4-\sqrt{x^2+y^2} \ dy\ dx\\ \end{align}\]We hit a roadblock here, and we notice the \(x^2+y^2\) in the expression, which tells us that cylindrical coordinates may work better:

\[\begin{align} m &= \int_0^{2\pi} \int_0^4 \int_0^{4-r} r \ dz\ dr\ d\theta\\ &= \int_0^{2\pi} \int_0^4 4r-r^2\ dr\ d\theta\\ &= \int_0^{2\pi} \dfrac{32}{3}\ d\theta = \dfrac{64\pi}{3} \end{align}\]Next, apply the formulas to find the center of mass, noting again that \(x=r\cos\theta\) and \(y=r\sin\theta\):

\[\begin{align} \overline{x} &= \dfrac{3}{64\pi} \int_{0}^{2\pi} \int_0^4 \int_0^{4-r} r^2 \cos \theta \ dz\ dr\ d\theta\\ &= \dfrac{3}{64\pi} \int_{0}^{2\pi} \int_0^4 (4r^2-r^3) \cos \theta \ dr\ d\theta\\ &= \dfrac{3}{64\pi} \int_{0}^{2\pi} \left(\dfrac{4}{3}r^3-\dfrac{1}{4}r^4\right)\bigg|_0^4 \cos \theta \ d\theta\\ &= \dfrac{3}{64\pi} \int_{0}^{2\pi} \dfrac{64}{3} \cos \theta \ d\theta\\ &= \dfrac{1}{\pi} \int_{0}^{2\pi} \cos \theta \ d\theta = 0\\ \textrm{Similarly, } \overline{y} &= 0\\ \overline{z} &= \dfrac{3}{64\pi} \int_{0}^{2\pi} \int_0^4 \int_0^{4-r} zr \ dz\ dr\ d\theta\\ &= \dfrac{3}{64\pi} \int_{0}^{2\pi} \int_0^4 \dfrac{1}{2}(4-r)^2r\ dr\ d\theta\\ &= \dfrac{3}{128\pi} \int_{0}^{2\pi} \int_0^4 16r-8r^2+r^3\ dr\ d\theta\\ &= \dfrac{3}{128\pi} \int_{0}^{2\pi} \dfrac{64}{3}\ d\theta = 1\\ \end{align}\]Therefore, the center of mass of this cone is located at

\[\ans{(\overline{x},\overline{y},\overline{z}) = \left(0,0,1\right)}\]Consider the hemisphere \(D\) with radius 4, centered at the origin, and its base on the xy plane. If the density of this hemisphere is \(\rho(\rho,\phi,\theta) = 2-\dfrac{\rho}{4}\) (note the abuse of notation, with \(\rho\) meaning both the spherical radius and the density function), find its center of mass.

Since the body is part of a sphere, we'll use spherical coordinates. If that wasn't enough of a tipoff, the density function was given in terms of spherical coordinates. To use spherical coordinates, we need to recall that \(x=\rho\ \sin\phi\ \cos\theta\), \(y=\rho\ \sin\phi\ \sin\theta\), and \(z=\rho\ \cos\phi\).

\[\begin{align} m &= \iiint_D \rho(\rho,\phi,\theta)\ \rho^2\ \sin\phi\ d\rho\ d\phi\ d\theta\\ &= \int_{0}^{2\pi} \int_{0}^{\pi/2} \int_0^4 \left(2-\dfrac{\rho}{4}\right) \rho^2\ \sin\phi\ d\rho\ d\phi\ d\theta\\ &= \int_{0}^{2\pi} \int_{0}^{\pi/2} \dfrac{80}{3} \sin\phi\ d\phi\ d\theta\\ &= \int_{0}^{2\pi} -\dfrac{80}{3} \cos\phi\bigg|_0^{\pi/2}\ d\theta\\ &= \int_{0}^{2\pi} \dfrac{80}{3} \ d\theta = \dfrac{160\pi}{3} \end{align}\]You should check this, but it turns out that \(\overline{x} = \overline{y} = 0\), which shouldn't be surprising, since the hemisphere is centered at the origin.

\[\begin{align} \overline{z} &= \dfrac{3}{160\pi} \int_{0}^{2\pi} \int_{0}^{\pi/2} \int_0^4 \rho\ \cos\phi\ \left(2-\dfrac{\rho}{4}\right) \rho^2\ \sin\phi\ d\rho\ d\phi\ d\theta\\ &= \dfrac{3}{160\pi} \int_{0}^{2\pi} \int_{0}^{\pi/2} \int_0^4 \left(\rho^3-\dfrac{1}{8}\rho^4\right)\ \sin 2\phi\ d\rho\ d\phi\ d\theta\\ &= \dfrac{3}{160\pi} \int_{0}^{2\pi} \int_{0}^{\pi/2} \dfrac{192}{5} \sin 2\phi\ d\phi\ d\theta\\ &= \dfrac{18}{50\pi} \int_{0}^{2\pi} \cos 2\phi\bigg|_0^{\pi/2}\ d\theta\\ &= \dfrac{18}{50\pi} \int_{0}^{2\pi} 2\ d\theta = \dfrac{36}{25} \end{align}\]Therefore, the center of mass of this hemisphere is located at

\[\ans{(\overline{x},\overline{y},\overline{z}) = \left(0,0,\dfrac{36}{25}\right)}\]- Find the center of mass of the tetrahedron in the first octant bounded by \(z=1-x-y\) and the coordinate planes if the density function is \(\rho(x,y,z)=1\).
- Consider the rectangular prism \(D=\{(x,y,z)\ :\ 0 \leq x \leq 4, 0 \leq y \leq 1, 0 \leq z \leq 1\}\). If the density of this prism is \(\rho(x,y,z) = 1+\dfrac{x}{2}\), find its center of mass.
- Find the center of mass of the solid bounded by the upper half of the sphere \(\rho=6\) and \(z=0\) with \(\rho(\rho,\phi,\theta)=1+\dfrac{\rho}{4}\).
- Find the center of mass of the prism formed by \(z=x\), \(x=1\), \(y=4\), and the coordinate planes, if \(\rho(x,y,z) = 2+y\).

\((\overline{x},\overline{y},\overline{z}) = \left(\dfrac{1}{4},\dfrac{1}{4},\dfrac{1}{4}\right)\)

\((\overline{x},\overline{y},\overline{z}) = \left(\dfrac{7}{3},\dfrac{1}{2},\dfrac{1}{2}\right)\)

\((\overline{x},\overline{y},\overline{z}) = \left(0,0,\dfrac{198}{85}\right)\)

\((\overline{x},\overline{y},\overline{z}) = \left(\dfrac{2}{3},\dfrac{7}{3},\dfrac{1}{3}\right)\)