
3.6: Application: Center of Mass

Recall from single-variable calculus that for a region $$R = {(x, y) : a ≤ x ≤ b,0 ≤ y ≤ f (x)}$$ in $$\mathbb{R}^2$$ that represents a thin, flat plate (Figure 3.6.1), where $$f (x)$$ is a continuous function on $$[a,b]$$, the center of mass of $$R$$ has coordinates $$(\bar x, \bar y)$$ given by

$\nonumber \bar x =\dfrac{M_y}{M} \text{ and }\bar y = \dfrac{M_x}{M}$

Figure 3.6.1 Center of mass of $$R$$

where

$M_x = \int_a^b \dfrac{(f(x))^2}{2}dx,\quad M_y = \int_a^b x f (x)\,dx, \quad M=\int_a^b f (x)\,dx \label{Eq3.27}$

assuming that $$R$$ has uniform density, i.e the mass of $$R$$ is uniformly distributed over the region. In this case the area $$M$$ of the region is considered the mass of $$R$$ (the density is constant, and taken as 1 for simplicity).

In the general case where the density of a region (or lamina) $$R$$ is a continuous function $$δ = δ(x, y)$$ of the coordinates $$(x, y)$$ of points inside $$R$$ (where $$R$$ can be any region in $$\mathbb{R}^2$$ ) the coordinates $$(\bar x,\bar y)$$ of the center of mass of $$R$$ are given by

$\bar x =\dfrac{M_y}{M} \text{ and }\bar y = \dfrac{M_x}{M}\label{Eq3.28}$

where

$M_y = \iint\limits_R xδ(x, y)\,d A ,\quad M_x = \iint\limits_R yδ(x, y)\,d A,\quad M=\iint\limits_R δ(x, y)\,d A ,\label{Eq3.29}$

The quantities $$M_x \text{ and }M_y$$ are called the moments (or first moments) of the region $$R$$ about the $$x$$-axis and $$y$$-axis, respectively. The quantity $$M$$ is the mass of the region $$R$$. To see this, think of taking a small rectangle inside $$R$$ with dimensions $$∆x \text{ and }∆y$$ close to 0. The mass of that rectangle is approximately $$δ(x_∗, y_∗)∆x∆y$$, for some point $$(x_∗, y_∗)$$ in that rectangle. Then the mass of $$R$$ is the limit of the sums of the masses of all such rectangles inside $$R$$ as the diagonals of the rectangles approach 0, which is the double integral $$\iint\limits_R δ(x, y)\,d A$$.

Note that the formulas in Equation \ref{Eq3.27} represent a special case when $$δ(x, y) = 1$$throughout $$R$$ in the formulas in Equation \ref{Eq3.29}.

Example 3.13

Find the center of mass of the region $$R = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 2x^2 }$$, if the density function at $$(x, y) \text{ is }δ(x, y) = x+ y$$.

Solution:

The region $$R$$ is shown in Figure 3.6.2. We have

Figure 3.6.2

\nonumber \begin{align} M&=\iint\limits_R δ(x, y)\,d A \\ \nonumber &= \int_0^1 \int_0^{2x^2}(x+ y)\,d y\, dx \\ \nonumber &=\int_0^1 \left ( xy + \dfrac{y^2}{2} \Big |_{y=0}^{y=2x^2} \right ) \,dx \\ \nonumber &=\int_0^1 (2x^3 +2x^4 )\,dx \\ \nonumber &= \dfrac{x^4}{2}+\dfrac{2x^5}{5} \Big |_0^1 = \dfrac{9}{10}\\ \end{align}

and

$\nonumber \begin{split} M_x &= \iint\limits_R yδ(x, y)\,d A \\ \nonumber &=\int_0^1 \int_0^{2x^2} y(x+ y)\,d y\, dx \\ \nonumber &=\int_0^1 \left ( \dfrac{xy^2}{2}+\dfrac{y^3}{3} \Big |_{y=0}^{y=2x^2} \right )\,dx \\ \nonumber &=\int_0^1 (2x^5 + \dfrac{8x^6}{ 3} )\,dx \\ \nonumber &=\dfrac{x^6}{3} + \dfrac{8x^7}{21} \Big |_0^1 = \dfrac{5}{7} \\ \end{split} \qquad \nonumber \begin{split} M_y &= \iint\limits_R xδ(x, y)\,d A \\ \nonumber &=\int_0^1 \int_0^{2x^2} x(x+ y)\,d y\, dx \\ \nonumber &=\int_0^1 \left ( x^2y+\dfrac{xy^2}{2} \Big |_{y=0}^{y=2x^2} \right )\,dx \\ \nonumber &=\int_0^1 (2x^4 + 2x^5 )\,dx \\ &=\dfrac{2x^5}{5} + \dfrac{x^6}{3} \Big |_0^1 = \dfrac{11}{15} \\ \end{split}$

so the center of mass $$(\bar x,\bar y)$$ is given by

$\nonumber \bar x =\dfrac{M_y}{M} = \dfrac{11/15}{9/10} = \dfrac{22}{27},\quad \bar y = \dfrac{M_x}{M} = \dfrac{5/7}{9/10}=\dfrac{50}{63}$

Note how this center of mass is a little further towards the upper corner of the region $$R$$ than when the density is uniform (it is easy to use the formulas in Equation \ref{Eq3.27} to show that $$(\bar x,\bar y) = \left ( \dfrac{3}{ 4} , \dfrac{3}{ 5} \right )$$ in that case). This makes sense since the density function $$δ(x, y) = x + y$$ increases as $$(x, y)$$ approaches that upper corner, where there is quite a bit of area.

In the special case where the density function $$δ(x, y)$$ is a constant function on the region $$R$$, the center of mass $$(\bar x,\bar y)$$ is called the centroid of $$R$$.

The formulas for the center of mass of a region in $$\mathbb{R}^2$$ can be generalized to a solid $$S$$ in $$\mathbb{R}^ 3$$ . Let $$S$$ be a solid with a continuous mass density function $$δ(x, y, z)$$ at any point $$(x, y, z)$$ in $$S$$. Then the center of mass of $$S$$ has coordinates $$(\bar x,\bar y,\bar z)$$, where

$\bar x = \dfrac{M_{yz}}{M},\quad \bar y = \dfrac{M_{xz}}{M},\quad \bar z= \dfrac{M_{xy}}{M},\label{Eq3.30}$

where

$M_{yz} = \iiint\limits_S xδ(x, y, z)\,dV, \quad M_{xz} = \iiint\limits_S yδ(x, y, z)\,dV,\quad M_{xy} = \iiint\limits_S zδ(x, y, z)\,dV ,\label{Eq3.31}$

$M = \iiint\limits_S δ(x, y, z)\,dV .\label{Eq3.32}$

In this case, $$M_{yz}, M_{xz}\text{ and }M_{x y}$$ are called the moments (or first moments) of $$S$$ around the $$yz$$-plane, $$xz$$-plane and $$x y$$-plane, respectively. Also, $$M$$ is the mass of $$S$$.

Example 3.14

Find the center of mass of the solid $$S = {(x, y, z) : z ≥ 0, x^2 + y^2 + z^2 ≤ a^2 }$$, if the density function at $$(x, y, z) \text{ is }δ(x, y, z) = 1$$.

Solution:

The solid $$S$$ is just the upper hemisphere inside the sphere of radius $$a$$ centered at the origin (see Figure 3.6.3). So since the density function is a constant and $$S$$ is symmetric about the $$z$$-axis, then it is clear that $$\bar x = 0 \text{ and }\bar y = 0$$, so we need only find $$\bar z$$. We have

$\nonumber M = \iiint\limits_S δ(x, y, z)\,dV = \iiint\limits_S 1dV = \text{ Volume}(S).$

Figure 3.6.3

But since the volume of $$S$$ is half the volume of the sphere of radius $$a$$, which we know by Example 3.12 is $$\dfrac{4\pi a}{3}$$, then $$M = \dfrac{2\pi a}{3}$$ . And

\nonumber \begin{align} M_{xy} &= \iiint\limits_S zδ(x, y, z)\,dV \\ \nonumber &=\iiint\limits_S z \,dV,\text{ which in spherical coordinates is} \\ \nonumber &=\int_0^{2\pi} \int_0^{\pi /2} \int_0^a (ρ \cos{φ})ρ^2 \sin{φ}\,dρ\, dφ\,dθ \\ \nonumber &=\int_0^{2\pi} \int_0^{\pi/2} \sin{φ} \cos{φ} \left ( \int_0^a ρ^3 \,dρ \right ) \,dφ\,dθ \\ \nonumber &= \int_0^{2\pi} \int_0^{\pi/2} \dfrac{a^4}{4} \sin{φ} \cos{φ}\,dφ\,dθ \\ \nonumber M_{xy}&=\int_0^{2\pi} \int_0^{\pi/2} \dfrac{a^4}{8} \sin{2φ}\,dφ\,dθ \quad (\text{since }\sin{2φ} = 2\sin{φ} \cos{φ}) \\ \nonumber &= \int_0^{2\pi} \left ( -\dfrac{a^4}{16} \cos{2φ} \Big |_{φ=0}^{ φ=\pi/2}\right ) \,dθ \\ \nonumber &=\int_0^{2\pi} \dfrac{a^4}{8}\,dθ \\ &=\dfrac{\pi a^4}{4}, \\ \end{align}

so

$\nonumber \bar z = \dfrac{M_{xy}}{M} = \dfrac{\dfrac{\pi a^4}{4}}{\dfrac{2\pi a^3}{3}}=\dfrac{3a}{8}.$

Thus, the center of mass of $$S$$ is $$(\bar x,\bar y,\bar z) = \left ( 0,0, \dfrac{3a}{8} \right ).$$