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Section 1.9 Applications of multiple integrals

Subsection 1.9.1 Center of mass

Definition 1.9.1. Center of mass.

Let \(\mathcal{R}\subseteq \R^n\) (\(n=2,3\)) represent a solid object, and suppose the function \(\delta\colon\mathcal{R}\rightarrow \R\) defines the mass density per unit area (\(n=2\)) or unit volume (\(n=3\)) at points in \(\mathcal{R}\text{.}\) We define the mass, first moments, and center of mass of \(\mathcal{R}\) as follows.

\begin{align*} n =2 \\ \text{Mass: } M\amp=\iint\limits_\mathcal{R}\delta(x,y)\, dA \\ \text{First moments}\colon M_{y}\amp=\iint\limits_\mathcal{R}x\delta(x,y)\, dA \amp M_{x}\amp=\iint\limits_\mathcal{R}y\delta(x,y)\, dA \\ \text{Center of mass}\colon (\overline{x},\overline{y})\amp=\frac{1}{M}(M_y, M_x) \\ n =3 \\ \text{Mass: } M\amp=\iiint\limits_\mathcal{R}\delta(x,y,z)\, dV \\ \text{First moments}\colon M_{yz}\amp=\iiint\limits_\mathcal{R}x\delta(x,y,z)\, dV \amp M_{xz}\amp=\iiint\limits_\mathcal{R}y\delta(x,y,z)\, dV \amp M_{xy}\amp=\iiint\limits_\mathcal{R}z\delta(x,y,z)\, dV\\ \text{Center of mass}\colon (\overline{x},\overline{y},\overline{z})\amp=\frac{1}{M}(M_{yz}, M_{xz}, M_{xy}) \text{.} \end{align*}
When the mass density function \(\delta\) is a constant function (i.e., \(\delta(x_1,x_2,\dots, x_n)=c\) for fixed \(c\in \R\)), we call the center of mass of \(\mathcal{R}\) its centroid. Equivalently, the centroid of \(\mathcal{R}\) is its center of mass, assuming \(\delta=1\) everywhere.

Example 1.9.2. Centroids of triangles.

Let \(T\) be the solid triangular region with vertices \((0,0),(1,0),(0,1)\text{.}\)

  1. Compute the center of mass of \(T\) when the density function \(\delta(x,y)=c\) is constant. (In other words, compute the centroid of \(T\)).

  2. Compute the center of mass of \(T\) when the density function is given by \(\delta(x,y)=x+y\text{.}\)

  3. Let \(T'\) be the solid triangular region with vertices \((0,0),(a,b),(c,d)\text{.}\) Give a formula for the centroid of \(T'\) in terms of \(a,b,c,d\text{.}\)

Solution.

  1. We compute

    \begin{align*} M \amp = \iint\limits_Tc\, dA \\ \amp =c\operatorname{area} T=\frac{c}{2}\\ M_y \amp=\iint\limits_T cx\, dA \\ \amp=c\int_0^1\int_0^{1-x}x\, dy\, dx \\ \amp = c\left(\frac{1}{2}x-\frac{1}{3}x^3\right)\Bigr\vert_0^1\\ \amp = \frac{c}{6}\\ M_x \amp =\iint\limits_T cy\, dA \\ \amp =c\int_0^1\int_0^{1-y}y\, dx\, dy\\ \amp =\frac{c}{6} \\ (\overline{x},\overline{y}) \amp=(M_y/M, M_x/M) \\ \amp = (1/3,1/3))\text{.} \end{align*}

    Note: intuitively it is no surprise that the center of mass of \(T\) lies along the line \(y=x\text{,}\) since this is an axis of symmetry of \(T\text{.}\) To make this symmetry argument rigorous, however, requires some finessing.

  2. We compute

    \begin{align*} M \amp = \iint\limits_T x+y\, dA \\ \amp=\int_0^1\int_0^{1-x}x+y\, dy\, dx \\ \amp =\int_0^1x(1-x)+\frac{1}{2}(1-x)^2\, dx\\ \amp = \frac{1}{3}\\ M_y \amp =\iint\limits_T x(x+y)\, dA\\ \amp = \int_0^1\int_0^{1-x}x^2+xy\, dy\, dx\\ \amp = \int_0^1x^2(1-x)+\frac{1}{2}x(1-x)^2\, dx\\ \amp = \frac{1}{8}\\ M_x \amp = \iint\limits_T y(x+y)\, dA \\ \amp =\int_0^1\int_0^{1-y}xy+y^2\, dx\, dy\\ \amp =\frac{1}{8}\\ (\overline{x},\overline{y}) \amp=(M_y/M, M_x/M) \\ \amp = (3/8, 3/8)\text{.} \end{align*}

    Again, it is no surprise that the center of mass lies along the line \(y=x\text{,}\) since both \(T\) and the density function \(\delta(x,y)=x+y\) are symmetric with respect to this line. In contrast to the previous computation, the center of mass is further from the origin. This makes sense in terms of our “balancing point” interpretation of center of mass: in this case the greater amount of mass in the upper right of the triangle moves the balancing point up and to the right.

  3. The transformation \(G(u,v)=(au+cv, bu+dv)\) maps the triangle \(T\) with vertices \((0,0),(1,0), (0,1)\) in the \(uv\)-plane to the given triangle \(T'\) in the \(xy\)-plane. We can use substitution to reduce the computation for \(T'\) to that of \(T\text{.}\) Let \(M, M_y, M_x\) be the mass and moments of \(T\text{,}\) and \(M', M_y', M_x'\) the mass and moments of \(T'\text{.}\)

    \begin{align*} M'\amp=\iint\limits_{T'}1\, dA \\ \amp =\iint\limits_T \abs{J(G)}\, dA \\ \amp =\iint\limits_T\abs{ad-bc}\, dA\\ \amp =\abs{ad-bc}M\\ M_y' \amp=\iint\limits_{T'}x\, dA \\ \amp =\iint\limits_T (au+cv) \abs{ad-bc}\, dA\\ \amp =\abs{ad-bc}(a\iint\limits_Tu\, dA+c\iint\limits_T v\, dA)\\ \amp =\abs{ad-bc}(aM_y+cM_x)\\ M_x' \amp=\iint\limits_{T'}y\, dA \\ \amp =\iint\limits_T (bu+dv) \abs{ad-bc}\, dA\\ \amp =\abs{ad-bc}(b\iint\limits_Tu\, dA+d\iint\limits_T v\, dA)\\ \amp =\abs{ad-bc}(bM_y+dM_x)\\ (\overline{x}',\overline{y}') \amp = (M_y'/M', M_x'/M') \\ \amp = \frac{1}{M}(aM_y+cM_x, bM_y+dM_x)\\ \amp = \frac{1}{3}(a+c,b+d)\text{.} \end{align*}

    Note: looking closely at the penultimate line, we see that

    \begin{equation*} (\overline{x}',\overline{y}')=G(\overline{x}, \overline{y})\text{.} \end{equation*}

    In other words, the center of mass of \(T'=G(T)\) is the result of applying \(G\) to the center of mass of \(T\text{!}\) This is a consequence of the fact that \(G\) is a linear transformation.

Example 1.9.3. Centroid of hemisphere.

Let \(D\) be the solid hemisphere \(x^2+y^2+z^2\leq 4\text{,}\) \(z\geq 0\text{.}\) Find the center of mass of \(D\) assuming the mass density function is given by \(\delta(x,y,z)=z\text{.}\)

Solution.

It is inuitively clear from the symmetry of \(D\text{,}\) as well as the fact that \(\delta(x,y,z)=z\) does not depend on \(x\) or \(y\text{,}\) that \(M_{yz}=M_{xz}=0\text{.}\) In more detail let \(D=D_1\cup D_2\) where \(D_1\) is the right half of \(D\) (i.e., where \(x\geq 0\)), and \(D_2\) is the left half. Let \(f(x,y,z)=x\delta(x,y,z)=xz\text{.}\) The transformation \(G(x,y,z)=(-x,y,z)\) maps \(D_1\) onto \(D_2\text{.}\) By substitution we have

\begin{align*} \iiint_{D_2}f(x,y,z)\, dV\amp =\iiint_{D_1}f(-x,y,z)\abs{J(G)}\, dV\\ \amp =\iiint_{D_1}-xz\, dV\\ \amp =-\iiint_{D_1}f(x,y,z)\, dV\text{,} \end{align*}

and thus

\begin{align*} M_{yz} \amp=\iiint_D xz\, dV \\ \amp=\iiint_{D_1}xz\, dV+\iiint_{D_2}xz\, dV \\ \amp =\iiint_{D_1}xz\, dV-\iiint_{D_1}xz\, dV=0\text{.} \end{align*}

A similar argument works to show \(M_x=0\text{.}\)

To compute the remaining ingredients we will use spherical coordinates. Observe that the solid hemisphere \(D\) has spherical description

\begin{equation*} \{(\rho, \phi,\theta)\colon 0\leq \theta\leq 2\pi, 0\leq \phi\leq \pi/2, 0\leq \rho\leq 2\}\text{.} \end{equation*}

Now compute:

\begin{align*} M \amp=\iiint_D \delta(x,y,z)\, dV \\ \amp =\iiint_{D}z\, dV\\ \amp =\int_0^{2\pi}\int_0^{\pi/2}\int_0^2\rho\cos\phi\cdot \rho^2\sin\phi \amp \text{(spher. subst.)}\\ \amp = 2\pi \int_0^{\pi/2}\cos\phi\sin\phi\, d\phi\int_0^2\rho^3\, d\rho\\ \amp =\frac{2}{8}\pi(\sin^2\phi)\vert_0^{\pi/2}(\rho^4)\vert_0^{2}\\ \amp =\frac{1}{4}\pi 2^4=4\pi\\ M_{xy} \amp =\iiint_D z\delta(x,y,z)\, dV \\ \amp =\iiint_{D}z^2\, dV\\ \amp =\int_0^{2\pi}\int_0^{\pi/2}\int_0^2\rho^2\cos^2\phi\cdot \rho^2\sin\phi \amp \text{(spher. subst.)}\\ \amp = 2\pi \int_0^{\pi/2}\cos^2\phi\sin\phi\, d\phi\int_0^2\rho^4\, d\rho\\ \amp =\frac{2}{15}\pi(-\cos^3\phi)\vert_0^{\pi/2}(\rho^5)\vert_0^{2}\\ \amp =\frac{2}{15}\pi 2^5=\frac{64}{15}\pi\\ (\overline{x},\overline{y},\overline{z}) \amp = (0,0,M_{xy}/M)\\ \amp = \left(0,0,\frac{16}{15}\right)\text{.} \end{align*}

Subsection 1.9.2 Probability and density functions

Definition 1.9.4. Random variable.

Let \(S\) be any set, which we will consider to be a sample space. A random variable is a function

\begin{align*} X\colon S \amp\rightarrow \R \\ s\amp \mapsto X(s)\text{.} \end{align*}

Definition 1.9.5.

Let \(X, Y\) be random variables defined on the sample space \(S\text{.}\) A joint probability density function for \(X\) and \(Y\) is a function \(f\colon \R^2\rightarrow \R\) satisfying the following properties.

  1. \(\displaystyle f(x,y)\geq 0\)

  2. \(\displaystyle \int_{-\infty}^\infty\int_{-\infty}^\infty f(x,y)\, dx\, dy=1\)

  3. \(\operatorname{Prob}\left( \{s\in S\colon (X(s),Y(s))\in \mathcal{R}\} \right)=\iint\limits_\mathcal{R}f(x,y)\, dA\) for any \(\mathcal{R}\subseteq \R^2\) over which \(f\) is integrable.

Example 1.9.6. Joint probability density function.

Verify that the function

\begin{equation*} f(x,y)=\begin{cases} 6x^2y\amp (x,y)\in [0,1]\times [0,1]\\ 0\amp (x,y)\notin [0,1]\times [0,1] \end{cases} \end{equation*}

is a joint probability distribution function.

Solution.

Let \(R=[0,1]\times [0,1]\text{.}\) It is clear that \(f(x,y)\geq 0\) for all \((x,y)\text{.}\) Furthermore, we have

\begin{align*} \iint\limits_{\R^2}f(x,y)\, dA \amp = \iint\limits_{R}f(x,y)\, dA \amp ((x,y)\notin R\implies f(x,y)=0)\\ \amp=\int_0^1\int_0^16x^2y\, dy\, dx \\ \amp =6\int_0^1x^2\, dx\int_0^1 y\, dy\\ \amp =6(1/3)(1/2)=1\text{.} \end{align*}

Definition 1.9.7.

Let \(\mathcal{R}\subseteq\R^2\) have area \(A\text{.}\) The random variables \(X, Y\) are uniformly distributed over \(\mathcal{R}\) if their joint probabiltity disribution function is given by

\begin{equation*} f(x,y)=\begin{cases} \frac{1}{A}\amp \text{if } (x,y)\in \mathcal{R}\\ 0 \amp \text{if } (x,y)\not\in \mathcal{R} \end{cases}. \end{equation*}

Definition 1.9.8. Expected value.

Assume the random variables \(X,Y\) on the sample set \(S\) have joint probability density function \(f(x,y)\text{.}\) We define the expected values of \(X\) and \(Y\text{,}\) denoted \(\mu_X\) and \(\mu_Y\text{,}\) as

\begin{align*} \mu_X\amp= \int_{-\infty}^\infty\int_{-\infty}^\infty x\, f(x,y)\, dx\, dy\\ \mu_Y\amp= \int_{-\infty}^\infty\int_{-\infty}^\infty y\, f(x,y)\, dx\, dy \end{align*}

Example 1.9.9. Joint probability density function.

Assume random variables \(X\) and \(Y\) have joint probability density function

\begin{equation*} f(x,y)=\begin{cases} 6x^2y\amp (x,y)\in [0,1]\times [0,1]\\ 0\amp (x,y)\notin [0,1]\times [0,1] \end{cases}\text{.} \end{equation*}

  1. Compute \(\mu_X\) and \(\mu_Y\) for any two random variables \(X\) and \(Y\) with joint probability distribution function \(f\text{.}\)

  2. Compute the probability that \(X\geq Y\text{.}\)

Solution.

  1. We compute:

    \begin{align*} \mu_X \amp =\iint\limits_R xf(x,y)\, dA \\ \amp = \int_0^1\int_0^16x^3y\, dx\, dy \\ \amp = 6\int_0^1x^3\, dx\int_0^1y\, dy\\ \amp =6(1/4)(1/2)=3/4 \\ \mu_Y \amp =\iint\limits_R yf(x,y)\, dA \\ \amp = \int_0^1\int_0^16x^2y^2\, dx\, dy \\ \amp = 6\int_0^1x^2\, dx\int_0^1y^2\, dy\\ \amp =6(1/3)(1/3)=2/3\text{.} \end{align*}

  2. Let \(\mathcal{R}=\{(x,y)\in [0,1]\times[0,1]\colon x\geq y\}\text{.}\) We have

    \begin{align*} \operatorname{Prob}(\{s\in S\colon X(s)\geq Y(s)\}) \amp= \operatorname{Prob}(\{s\in S\colon (X(s),Y(s))\in \mathcal{R}\}) \\ \amp = \iint\limits_\mathcal{R}f(x,y)\, dA \amp (\knowl{./knowl/d_probability_functions.html}{\text{1.9.5}}) \\ \amp =\int_0^1\int_0^{x}6x^2y\, dy\, dx \\ \amp = \int_0^13x^4\\ \amp =\frac{3}{5}\text{.} \end{align*}