8.5 Applications in Physics and Statistics

This chapter concludes with a few applications showing how some familiar discrete sums can be replaced by integrals, which are essentially continuous sums.

Center of Gravity

Suppose a thin uniform rod has $n>1$ masses $m_1,\ldots,m_n$ attached, with $m_1$ and $m_n$ at the ends. The center of gravity of the masses is the point where—due to the Earth’s gravity—the rod would be balanced if a fulcrum were placed there (see Figure 8.5.1(a)). Imagine the rod as part of the $x$-axis and the weights as point masses—with each mass $m_k$ at $x_k$—and let the center of gravity be at $\bar{x}$, as in Figure 8.5.1(b).

tikzpicture[>=latex,every node/.style=font=] [black] (-0.3,-0.3) -- (0,-0.035) -- (0.3,-0.3) -- cycle; [linecolor,line width=1.5pt] (-2,0) -- (2,0); (-2,0) circle (4pt); (-1,0) circle (4pt); (1,0) circle (4pt); (2,0) circle (4pt); [above] at (-2,0.2) m_1; [above] at (-1,0.2) m_2; [above] at (0,0.2) ; [above] at (1,0.2) m_n-1; [above] at (2,0.2) m_n; tikzpicture — Figure 8.5.1 Center of gravity $\bar{x}$ for masses $m_1,\ldots,m_n$

tikzpicture[>=latex,every node/.style=font=] [->,black!60,line width=1pt,anchor=base] (-1,0) -- (5,0) node[right] x node[black,shift=(0,-0.4)] at (0,0) x_1 node[black,shift=(0,-0.4)] at (1,0) x_2 node[black,shift=(0,-0.4)] at (2,0) x node[black,shift=(0,-0.4)] at (3,0) x_n-1 node[black,shift=(0,-0.4)] at (4,0) x_n; [black!60] (2,0.1) -- (2,-0.1); [linecolor,line width=1.5pt] (0,0) -- (4,0); (0,0) circle (2.5pt); (1,0) circle (2.5pt); (3,0) circle (2.5pt); (4,0) circle (2.5pt); [above] at (0,0.1) m_1; [above] at (1,0.1) m_2; [above] at (2,0.1) ; [above] at (3,0.1) m_n-1; [above] at (4,0.1) m_n; tikzpicture — Figure 8.5.1 Center of gravity $\bar{x}$ for masses $m_1,\ldots,m_n$

The rod is balanced if the masses do not rotate the rod, i.e. the total torque is zero. Torque is defined here as force times position relative to $\bar{x}$. Each mass $m_k$ applies a force $m_kg$ to the rod—where $g$ is the (downward) acceleration due to the Earth’s gravity—at position $(x_x-\bar{x})$ relative to $\bar{x}$. The total torque is thus zero if

\[ (m_1g)\,(x_1-\bar{x}) ~+~ (m_2g)\,(x_2-\bar{x}) ~+~ \cdots ~+~ (m_ng)\,(x_n-\bar{x}) ~=~ 0 \]

so that solving for $\bar{x}$ yields:

\begin{equation} \bar{x} ~=~ \frac{m_1gx_1 + \cdots + m_ngx_n}{m_1g + \cdots + m_ng} ~=~ \frac{m_1x_1 + \cdots + m_nx_n}{m_1 + \cdots + m_n} ~=~ \frac{\sum_{k=1}^n \;m_kx_k}{\sum_{k=1}^n \;m_k} \tag{8.17}\end{equation}

Each quantity $m_kx_k$ is called the moment of the mass $m_k$. Thus, $\bar{x}$ is the sum of the moments divided by the total mass. This idea can be extended to regions in the $xy$-plane, using an integral of a continuum of moments instead of a finite sum. The center of gravity of a planar region is defined as the point such that any force along a line through that point produces no rotation of the region about that line.^[1] There should thus be zero torque in both the $x$ and $y$ directions, so the idea is to apply formula (8.17) in both directions to obtain the region’s center of gravity $(\bar{x},\bar{y})$. A region can be thought of as a lamina—a thin plate with uniform density. Take the area of the region as its mass, which makes sense given the uniform density. For the region between two curves $y=f_1(x)$ and $y=f_2(x)$ over $\ival{a}{b}$, with $f_1(x) \ge f_2(x)$, take a vertical slice of width $\dx$ at some $x$, as in Figure 8.5.2(a). By the same arguments used in Section 8.4, all the area from that strip comes from the rectangle of height $f_1(x)-f_2(x)$ and width $\dx$ (see the shaded rectangle in Figure 8.5.2(b)).

tikzpicture[>=latex,every node/.style=font=] [fillcolor] (1,0.8) -- (1,1.5) parabola (2,2) -- (2,0.6) -- cycle; (1,0.8) -- (1,1.5); (2,0.6) -- (2,2); [pattern=north west lines] (1.5,0.7) -- (1.5,1.6) -- (1.7,1.75) -- (1.7,0.65); [<->,black!60,line width=1pt,anchor=base] (0,2.1) node[above] y |- (3,0) node[right] x node[black,shift=(0,-0.4)] at (0,0) 0 node[black,shift=(0,-0.4)] at (1,0) a node[black,shift=(0,-0.4)] at (1.5,0) x node[black,shift=(0,-0.4)] at (2,0) b; (1.5,0.7) -- (1.5,1.6); (1.7,0.65) -- (1.7,1.75); [linecolor,line width=1.5pt] (1,1.5) parabola (2,2); [linecolor,line width=1.5pt] (1,0.8) -- (2,0.6); (1,1.5) circle (2.5pt); (2,2) circle (2.5pt); (1,0.8) circle (2.5pt); (2,0.6) circle (2.5pt); [left] at (1.7,1.9) y=f_1(x); [left] at (1.7,0.4) y=f_2(x); [black!60] (1,0.1) -- (1,-0.1); [black!60] (1.5,0.1) -- (1.5,-0.1); [black!60] (2,0.1) -- (2,-0.1); (-1.8,0) -- (4.5,0); tikzpicture — Figure 8.5.2 Center of gravity of a region

tikzpicture[>=latex,every node/.style=font=] [|<->|] (0.7,0.4) -- (0.7,1.5) node[fill=white,midway,xshift=-0.5cm] f_1(x)-f_2(x); [black!30] (1.05,0.4) -- (1.05,1.5) -- (1.5,1.5) -- (1.5,0.4) -- cycle; [dashed] (1.05,1.5) -- (1.5,1.5); [dashed] (1.05,0.4) -- (1.5,0.4); (1.05,0.4) -- (1.05,1.5); (1.5,0.2) -- (1.5,2.1); [<->,black!60,line width=1pt] (-1,2.1) node[above] y |- (3,0) node[right] x; [linecolor,line width=1.5pt] (1.05,1.5) -- (1.5,2.1) node[black,sloped,midway,above] ; [linecolor,line width=1.5pt] (1.05,0.4) -- (1.5,0.2); at (1.275,-0.2) ; [->|] (0.8,-0.2) -- (1.05,-0.2); [->|] (1.75,-0.2) -- (1.5,-0.2); [->|] (1.8,1.2) -- (1.8,1.5); [->|] (1.8,2.4) -- (1.8,2.1); at (1.8,1.8) ; (1.275,0.95) circle (1.5pt); [->] (2.1,0.95) -- (1.35,0.95); [right] at (2.1,0.95) (x+ 12 , 12(f_1(x)+f_2(x))); (-1,0) -- (6,0); tikzpicture — Figure 8.5.2 Center of gravity of a region

By the assumption of uniform density, the center of gravity of that rectangle is clearly its geometric center, whose coordinates are $\left(x+\frac{1}{2}\dx,\frac{1}{2}(f_1(x)+f_2(x))\right)$. The entire mass of the strip can be treated as if it is concentrated at that point. The moment $m_x$ of the strip about the $x$-axis is its mass times the position of its center of gravity relative to the $x$-axis (i.e. its $y$ coordinate):

\[ m_x ~=~ (f_1(x)-f_2(x))\,\dx\,\cdot\,(\tfrac{1}{2}\,(f_1(x)+f_2(x))) ~=~ \tfrac{1}{2}\,((f_1(x))^2 - (f_2(x))^2)\,\dx \]

Similarly the moment $m_y$ of the strip about the $y$-axis is its mass times the $x$ coordinate of its center of gravity:

\begin{align*} m_y ~&=~ (f_1(x)-f_2(x))\,\dx\,\cdot\,(x+\tfrac{1}{2}\dx) ~=~ x\,(f_1(x)-f_2(x))\,\dx ~+~ \tfrac{1}{2} (f_1(x)-f_2(x))\,(\dx)^2\\ ~&=~ x\,(f_1(x)-f_2(x))\,\dx \end{align*}

The moments $M_x$ and $M_y$ of the entire region about the $x$-axis and $y$-axis, respectively, are defined as the sum of the respective moments $m_x$ and $m_y$ of all strips over $\ival{a}{b}$:

\[ M_x \;=\; \int_a^b m_x ~=~ \int_a^b \tfrac{1}{2}\,((f_1(x))^2 - (f_2(x))^2)\;\dx \enskip\text{and}\enskip M_y \;=\; \int_a^b m_y ~=~ \int_a^b x\,(f_1(x)-f_2(x))\;\dx \]

Note in formula (8.17) that the denominator is the sum of all the masses in the system. For the region that total mass would simply be its area $M$:

\[ M ~=~ \int_a^b (f_1(x)-f_2(x))~\dx \]

Dividing the moments $M_x$ and $M_y$ by $M$ yields the formula for the center of gravity:

Theorem 8.12

The center of gravity $(\bar{x},\bar{y})$ of the region between the curves $y=f_1(x)$ and $y=f_2(x)$ over $\ival{a}{b}$, with $f_1(x) \ge f_2(x)$, is given by:

\begin{equation} \bar{x} ~=~ \frac{M_y}{M} ~=~ \frac{\displaystyle\int_a^b x\,(f_1(x)-f_2(x))~\dx}{\displaystyle\int_a^b (f_1(x)-f_2(x))~\dx} \quad\text{and}\quad \bar{y} ~=~ \frac{M_x}{M} ~=~ \frac{\displaystyle\int_a^b \tfrac{1}{2}\,((f_1(x))^2 - (f_2(x))^2)~\dx}{\displaystyle\int_a^b (f_1(x)-f_2(x))~\dx} \tag{8.18}\end{equation}

Example 8.24

Find the center of gravity of the region bounded by the curve $y=x^2$ and the $x$-axis for $0 \le x \le 1$. [r]

Solution: The region is shaded in the figure on the right. Using $y=f_1(x)=x^2$ and $y=f_2(x)=0$ in formula (8.18) yields

\begin{align*} M_x ~&=~ \int_0^1 \tfrac{1}{2}\,(f_1(x))^2~\dx ~=~ \int_0^1 \tfrac{1}{2}\,x^4~\dx ~=~ \tfrac{1}{10}\,x^5~\Biggr|_0^1 ~=~ \tfrac{1}{10}\\ M_y ~&=~ \int_0^1 x\,f_1(x)~\dx ~=~ \int_0^1 x^3~\dx ~=~ \tfrac{1}{4}\,x^4~\Biggr|_0^1 ~=~ \tfrac{1}{4}\\ M ~&=~ \int_0^1 f_1(x)~\dx ~=~ \int_0^1 x^2~\dx ~=~ \tfrac{1}{3}\,x^3~\Biggr|_0^1 ~=~ \tfrac{1}{3} \end{align*}

so that the center of gravity $(\bar{x},\bar{y})$ is:

\[ \bar{x} ~=~ \frac{M_y}{M} ~=~ \frac{1/4}{1/3} ~=~ \frac{3}{4} \quad\text{and}\quad \bar{y} ~=~ \frac{M_x}{M} ~=~ \frac{1/10}{1/3} ~=~ \frac{3}{10} \]

Example 8.25

[r] Find the center of gravity of the region bounded by the curves $y=x$ and $y=x^2$.

Solution: The region is shaded in the figure on the right. Using $y=f_1(x)=x$ and $y=f_2(x)=x^2$ in formula (8.18) yields

\begin{align*} M_x ~&=~ \int_0^1 \tfrac{1}{2}\,((f_1(x))^2-(f_2(x))^2)~\dx ~=~ \int_0^1 \tfrac{1}{2}\,(x^2-x^4)~\dx ~=~ \tfrac{1}{6}\,x^3 - \tfrac{1}{10}\,x^5~\Biggr|_0^1 ~=~ \tfrac{1}{15}\\ M_y ~&=~ \int_0^1 x\,(f_1(x)-f_2(x))~\dx ~=~ \int_0^1 (x^2 - x^3)~\dx ~=~ \tfrac{1}{3}\,x^3 - \tfrac{1}{4}\,x^4~\Biggr|_0^1 ~=~ \tfrac{1}{12}\\ M ~&=~ \int_0^1 (f_1(x) - f_2(x))~\dx ~=~ \int_0^1 (x-x^2)~\dx ~=~ \tfrac{1}{2}\,x^2 - \tfrac{1}{3}\,x^3~\Biggr|_0^1 ~=~ \tfrac{1}{6} \end{align*}

so that the center of gravity $(\bar{x},\bar{y})$ is:

\[ \bar{x} ~=~ \frac{M_y}{M} ~=~ \frac{1/12}{1/6} ~=~ \frac{1}{2} \quad\text{and}\quad \bar{y} ~=~ \frac{M_x}{M} ~=~ \frac{1/15}{1/6} ~=~ \frac{2}{5} \]

Work

Suppose that a constant force displaces an object along a line in the same direction in which the force is applied. The work done by the force is defined as the force times the displacement. For example, if the constant force $F$ moves an object from position $x=a$ to $x=b$ on the $x$-axis, as in Figure 8.5.3(a), then the work $W$ done by the force is:

\[ W ~=~ \text{force}\,\times\,\text{displacement} ~=~ \text{force}\,\times\,\text{(final position $-$ initial position)} ~=~ F\,\cdot\,(b-a) \]

tikzpicture[>=latex,every node/.style=font=] [fill=fillcolor] (0.6,0) rectangle (1.4,0.8); [fill=fillcolor,dashed] (3.6,0) rectangle (4.4,0.8); [->,black!60,line width=1pt,anchor=base] (-0.5,0) -- (5.5,0) node[right] x node[black,shift=(0,-0.4)] at (1,0) a node[black,shift=(0,-0.4)] at (4,0) b; [->,linecolor,line width=1.5pt] (1,0.4) -- (2.5,0.4) node[black,above,pos=0.8] F; (1,0.4) circle (2.5pt); [black!60] (1,0.1) -- (1,-0.1); [black!60] (4,0.1) -- (4,-0.1); tikzpicture — Figure 8.5.3 Work $W$ as the effect of a force $F$ displacing an object from $x=a$ to $x=b$

tikzpicture[>=latex,every node/.style=font=] (-0.7,0) -- (4.8,0); [dashed] (0,0) -- (2.5,0) node[midway,below] -- (2.5,1.5); [linecolor,line width=1.5pt] (0,0) -- (2.5,1.5); (2.3,0) -- (2.3,0.2) -- (2.5,0.2); [left] at (0,0) F(x); [right] at (2.5,1.5) F(x+ ); tikzpicture — Figure 8.5.3 Work $W$ as the effect of a force $F$ displacing an object from $x=a$ to $x=b$

Suppose now that the force $F$ is a function of position $x$ over $\ival{a}{b}$: $F=F(x)$. By the Microstraightness Property, over an infinitesimal interval $\ival{x}{x+\dx}$ the curve $y=F(x)$ is a straight line, as in Figure 8.5.3(b). How should the work $d\!W$ performed by $F$ over this infinitesimal interval be defined? After all, $F$ is not constant over $\ival{x}{x+\dx}$—it takes every value between $F(x)$ and $F(x+\dx)$. It is left as an exercise to show that any value in that range can be used—they all result in the same amount $F(x)\,\dx$ for the work performed.^[2]

For example, suppose you use the value halfway between $F(x)$ and $F(x+\dx)$ as the value of $F$: $\frac{1}{2}\,(F(x)+F(x+\dx))$. Then the work $d\!W$ as force times displacement is:

\begin{align*} d\!W ~&=~ \frac{1}{2}\,(F(x) + F(x+\dx))~\dx ~=~ \frac{1}{2}\,(F(x) + F(x)+ F'(x)\,\dx)~\dx\\[4pt] &=~ F(x)\,\dx ~+~ \frac{1}{2}\,F'(x)\,\cancelto{0}{(\dx)^2}\\ &=~ F(x)\,\dx \end{align*}

Define the total work $W$ over $\ival{a}{b}$ as the sum of all the $d\!W$:

Definition 8.4

The work performed by a force $F(x)$ in displacing an object along the $x$-axis from $x=a$ to $x=b$ is:

\begin{equation} W ~=~ \int_a^b d\!W ~=~ \int_a^b F(x)~\dx \tag{8.19}\end{equation}

Before continuing, some possible confusion needs to be cleared up. First, force is always a vector—it has both a magnitude and a direction. For the forces considered here, which act in a single dimension (e.g. along the $x$-axis), by convention the direction of the force is indicated by its sign: positive in the direction toward $+\infty$, negative in the direction toward $-\infty$. So a force of $3$ N acts in the opposite direction as a force of $-3$ N, but they have the same magnitude $\abs{3}=3$.

Second, work is not a vector—it is a scalar, meaning it has a magnitude but no direction. That magnitude can have any sign, though. Work is positive if the object is displaced in the same direction as the force, but is negative if the displacement is in the opposite direction of the force. For example, if you lift an object straight up from the ground, then you did positive work—the object moved in the same direction as the force you used. However, the force of gravity did negative work on the object as you lifted, since gravity works downward yet the object moved upward.

[r] Last, zero work is done by a force if no displacement in its direction occurs. In particular, forces acting perpendicular to the line of displacement perform no work. For example, consider an object of mass $m$ on a flat horizontal table top as in the figure on the right. If you push that object to the right with a force $F$ (performing positive work), then both the downward force of gravity $-mg$ and the upward normal force $N$ exerted by the table perform zero work on the object. The force of friction $F_{\mu}$ from the table surface does negative work, as it opposes the force $F$. As an another example, no work is performed by holding a $100$ lb object still and above the ground.

Example 8.26

[r] Hooke’s law states that a coiled spring has an elastic restoring force $F=-kx$, where $x$ is the displacement of the end of the spring from its equilibrium position as the spring is stretched or compressed, and $k>0$ is the spring constant—or stiffness coefficient—specific to the spring. This force always tries to restore the spring to its equilibrium position, and the law holds only for a limited range of $x$. For a spring laid horizontally imagine it lies on the $x$-axis with the equilibrium position at $x=0$, as in the figure on the right.

Find the spring constant $k$ if a force of 2 N stretches the spring by 4 cm.
Use part (a) to find the work performed by compressing the spring 3 cm.

Solution: (a) The force required to stretch the spring by an amount $x$ is $F=kx$, since that force must counter the restoring force. Thus, $k=\frac{F}{x}=\frac{2 \text{N}}{4 \text{cm}}=\frac{2 \text{N}}{0.04 \text{m}}= 50$ N/m.
(b) By part (a) the force required to compress the string to position $x$ is $F(x)=kx=50x$, since again it must counter the restoring force. Thus, since $3$ cm is $0.03$ m, the work $W$ performed is:

\[ W ~=~ \int_0^{-0.03} F(x)~\dx ~=~ \int_0^{-0.03} 50x~\dx ~=~ 25x^2~\Biggr|_0^{-0.03} ~=~ 25\,(-0.03)^2 ~-~ 0 ~=~ 0.0225~\text{Nm} \]

Probability

Suppose you flip two evenly balanced pennies and let $X$ be the number of heads in the result. Then $X$ is a discrete random variable—discrete because it can take only a discrete set of values (0, 1 and 2); random because its value is left to chance. The probability of a penny being flipped heads is $50\% = \frac{1}{2}$, i.e. that is its theoretical likelihood since heads and tails are equally likely. The sample space $S$ of all possible outcomes is the set $S = \lbrace TT, TH, HT, HH \rbrace$, where $H$ is heads and $T$ is tails (e.g. $HT$ means the first penny came up heads and the second came up tails). Figure 8.5.4(a) shows a bar chart of the probabilities—as numbers between 0 and 1—with $P(X=x)$ denoting the probability of the event that $X$ equals the number $x$. Notice that the sum of the probabilities is 1, and $P(X=x)=0$ if $x$ is not 0, 1, or 2.

tikzpicture[>=latex,every node/.style=font=] [fill=fillcolor,draw=linecolor,line width=1pt] (0.8,0) rectangle (1.2,0.6); [fill=fillcolor,draw=linecolor,line width=1pt] (1.8,0) rectangle (2.2,1.2); [fill=fillcolor,draw=linecolor,line width=1pt] (2.8,0) rectangle (3.2,0.6); [<->,black!60,line width=1pt,anchor=base] (0,1.8) node[above] P(X=x) |- (4,0) node[right] x node[black,shift=(0,-0.4)] at (1,0) 0 node[black,shift=(0,-0.4)] at (2,0) 1 node[black,shift=(0,-0.4)] at (3,0) 2; [black!60] (1,0) -- (1,-0.1); [black!60] (2,0) -- (2,-0.1); [black!60] (3,0) -- (3,-0.1); [black!60] (0.1,0.6) -- (-0.1,0.6) node[left,black] 14; [black!60] (0.1,1.2) -- (-0.1,1.2) node[left,black] 12; [rotate=90] at (-1,0.9) probability; tikzpicture — Figure 8.5.4 Probability: discrete vs continuous random variables

tikzpicture[>=latex,every node/.style=font=] [fillcolor] (0,0) -- plot[domain=0:2.5,samples=300] ( ,3* *exp(- * )) -- (2.5,0); [pattern=north west lines] (0.6,0) -- plot[domain=0.6:1.2,samples=100] ( ,3* *exp(- * )) -- (1.2,0); [linecolor,line width=1.5pt] plot[domain=0.0072:2.5,samples=300] ( ,3* *exp(- * )); [<->,black!60,line width=1pt,anchor=base] (0,1.8) node[above] f(x) |- (3,0) node[right] x node[black,shift=(0,-0.4)] at (0.6,0) a node[black,shift=(0,-0.4)] at (1.2,0) b; [black!60] (0.6,0) -- (0.6,-0.1); [black!60] (1.2,0) -- (1.2,-0.1); (-1.5,0) -- (4,0); tikzpicture — Figure 8.5.4 Probability: discrete vs continuous random variables

The idea behind a continuous random variable $X$ is to fill in those gaps between the bars in Figure 8.5.4(a), so that $X$ would represent a continuous quantity, e.g. time, distance, temperature. Rather than finding $P(X=x)$ you would find the probability that $X$ is in a continuum such as an interval, e.g. $P(a < X < b)$ (see Figure 8.5.4(b)).

Definition 8.5

For a continuous random variable $X$ define $P(X=x)=0$ for all real $x$, and define the probability density function for $X$ as a function $f(x) \ge 0$ such that:

$\displaystyle\int_{-\infty}^{\infty} f(x)~\dx ~=~ 1$
$P(a < X < b) ~=~ \displaystyle\int_a^b f(x)~\dx~~$ for all $a < b$ (including $a=-\infty$ and $b=\infty$)

Notice that since $P(X=a)=0$ then $P(a \le X < b) = P(a < X < b)$. In general, $<$ and $\le$ are interchangeable for events involving continuous random variables (as well as $>$ and $\ge$). In the remainder of this section it will be assumed that all random variables are continuous, for which the sample space is typically all of $\Reals$ or some interval, finite or infinite (e.g. $(0,\infty)$).

Example 8.27

Let $X$ be the lifetime—i.e. the time to failure—of an electronic component. If the average lifetime of the component is 700 days, then the probability density function $f(x)$ for the random variable $X$ is

\begin{equation} f(x) ~=~ \begin{cases} ~\lambda\,e^{-\lambda x}& \text{if $~x \ge 0$,}\\~0 & \text{if $~x<0$}\end{cases} \tag{8.20}\end{equation}

where $\lambda = \frac{1}{700}$ and $x$ is the number of days. In this case $X$ is said to have the exponential distribution with parameter $\lambda$. Find the probability that the lifetime of the component is:

between 600 and 800 days
greater than 700 days

Solution: (a) The probability is:

\[ P(600 < X < 800) ~=~ \int_{600}^{800} f(x)~\dx ~=~ \int_{600}^{800} \tfrac{1}{700}\,e^{-\frac{x}{700}}~\dx ~=~ -e^{-\frac{x}{700}}~\Biggr|_{600}^{800} ~=~ -e^{-\frac{800}{700}} ~+~ e^{-\frac{600}{700}} ~\approx~ 0.1055 \]

Thus, there is about a $10.55\%$ chance that the component’s lifetime will be between 600 and 800 days.

(b) The probability is:

\[ P(X > 700) ~=~ \int_{700}^{\infty} f(x)~\dx ~=~ \int_{700}^{\infty} \tfrac{1}{700}\,e^{-\frac{x}{700}}~\dx ~=~ -e^{-\frac{x}{700}}~\Biggr|_{700}^{\infty} ~=~ 0 ~+~ e^{-1} ~\approx~ 0.3679 \]

Exercises

For Exercises 1-3, find the center of gravity of the region bounded by the given curves over the given interval.

$y = x^3$ and $y = 0$ ; $0\le x\le 1$
$y= -x+1$ and $y = 0$ ; $0\le x\le 1$
$y = x^2$ and $y = x^3$ ; $0\le x\le 1$
Find the center of gravity of the region inside the circle $x^2+y^2=r^2$ and above the $x$-axis.
Find the center of gravity of the region inside the circle $x^2+y^2=r^2$ in the first quadrant.
Find the center of gravity of the region inside the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and above the $x$-axis.
Find the center of gravity of the region between the circle $x^2+y^2=4$ and the ellipse $\frac{x^2}{4}+y^2=1$ above the $x$-axis.
Would formula (8.18) for the center of gravity change if the mass of a region were proportional—but not equal—to its area, say, by a constant positive proportion $\delta \ne 1$? Explain.
If a spring requires 3 N of force to be compressed 5 cm, how much work would be performed in stretching the spring 8 cm?
The gravitational force $F(x)$ exerted by the Earth on an object of mass $m$ at a distance $x$ from the center of the Earth is

\[ F(x) ~=~ -\frac{mgr_e^2}{x^2} \]

where $r_e$ is the radius of the Earth. If the object is released from rest at a distance $r_o$ from the center of the Earth, find the work performed by gravity in bringing the object to the Earth’s surface.
Recall that the ideal gas law states that $PV=RT$, where $R$ is a constant, $P$ is the pressure, $V$ is the volume, and $T$ is the temperature. It can be shown that the work $W$ done by an ideal gas in expanding the volume from $V_a$ to $V_b$ is

\[ W ~=~ \int_{V_a}^{V_b} P~d\!V ~. \]

Calculate $W$.
Verify that $~\displaystyle\int_{-\infty}^{\infty} f(x)~\dx = 1~$ for the function $f(x)$ in formula (8.20) in Example 8.27 for all $\lambda > 0$.
Find $P(X < 300)$ in Example 8.27.
The distribution function $F(x)$ for a random variable $X$ is defined as $F(x) = P(X \le x)$ for all $x$. Show that $F'(x) = f(x)$, where $f(x)$ is the probability density function for $X$.

Formula (8.18) can be extended to regions over an infinite interval, provided the area is finite. Use that fact to find the center of gravity of the region between $y=e^{-x}$ and the $x$-axis for $0\le x<\infty$.
The expected value (or mean) $E\lbrack X\rbrack$ of a random variable $X$ with probability density function $f(x)$ is

\[ E\lbrack X\rbrack ~=~ \int_{-\infty}^{\infty} x\;f(x)~\dx ~. \]

Show that $E\lbrack X\rbrack = \frac{1}{\lambda}$ if $X$ has the exponential distribution with parameter $\lambda >0$.
Note: The expected value can be thought of as the weighted average of all possible values of $X$, with weights determined by probability. It is analogous to the idea of a center of gravity.
A random variable $X$ is said to have a normal distribution if its probability density function $f(x)$ is

\[ f(x) ~=~ \frac{1}{\sigma\,\sqrt{2\pi}}\,e^{\frac{(x-\mu)^2}{2\sigma^2}} \quad\text{for all $x$} \]

where $\sigma > 0$ and $\mu$ are constants. This is the famous “bell curve” in statistics.
1. Verify that $~\displaystyle\int_{-\infty}^{\infty} f(x)~\dx = 1$. (Hint: Use Example 6.25 and a substitution.)
2. Show that $E\lbrack X\rbrack = \mu$.
3. Use numerical integration to show that $P(-1 < X < 1)\;\approx\; 0.6827\;$ when $\mu=0$ and $\sigma=1$.
A random variable $X$ has the beta distribution if its probability density function $f(x)$ is

\[ f(x) ~=~ \begin{cases} ~\frac{1}{B(a,b)}\,x^{a-1}\,(1-x)^{b-1} & \text{if $~0\le x\le 1$}\\ ~0 & \text{elsewhere}\end{cases} \]

for positive constants $a$ and $b$, where $B(a,b)$ is the Beta function. Show that $E\lbrack X\rbrack = \frac{a}{a+b}$.
Show that any value between $F(x)$ and $F(x+\dx)$ for the force over $\ival{x}{x+\dx}$ gives the same formula $d\!W = F(x)\,\,dx$ for the work performed over that interval. (Hint: Consider $F(x+\alpha\,\dx)$ for $0\le\alpha\le 1$.)
A drop of water of mass $M$ is released from rest at a height sufficient for the drop to evaporate completely, losing mass $m$ each second (i.e. at a constant rate). Ignoring air resistance, show that the work performed by gravity on the drop up to complete evaporation is $\frac{g^2 M^2}{6 m^2}$.
This exercise is related to Einstein’s famous law $E = mc^2$. The relativistic momentum $p$ of a particle of mass $m$ moving at a speed $v$ along a straight line (say, the $x$-axis) is

\[p ~=~ \dfrac{mv}{\sqrt{1 - \frac{v^2}{c^2}}} ~,\]

where $c$ is the speed of light. The relativistic force on the particle along that line is

\[F ~=~ \dfrac{d\!p}{\dt} ~,\]

which is the same formula as Newton’s Second Law of motion in classical mechanics. Assume that the particle starts at rest at position $x_1$ and ends at position $x_2$ along the $x$-axis. The work done by the force $F$ on the particle is:

\[W ~=~ \displaystyle\int_{x_1}^{x_2}~F~\dx ~=~ \displaystyle\int_{x_1}^{x_2}~\dfrac{d\!p}{\dt}~\dx\]
1. Show that
  
  \[ \dfrac{d\!p}{\dv} ~=~ \dfrac{m}{\left(1 - \frac{v^2}{c^2}\right)^{3/2}} ~. \]
2. Use the Chain Rule formula
  
  \[ \dfrac{d\!p}{\dt} ~=~ \dfrac{d\!p}{\dv}\;\dfrac{\dv}{\dx}\;\dfrac{\dx}{\dt} \]
  
  to show that
  
  \[ F\;\dx ~=~ v\;\dfrac{d\!p}{\dv}\;\dv ~. \]
3. Use parts (a) and (b) to show that
  
  \[ W ~=~ \displaystyle\int_{0}^{v}~\dfrac{d\!p}{\dv}\;v~\dv ~=~ \displaystyle\int_{0}^{v}~\dfrac{mv}{\left(1 - \frac{v^2}{c^2}\right)^{3/2}}~\dv ~~. \]
4. Use part (c) to show that
  
  \[ W ~=~ \dfrac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}} \;-\;mc^2 ~. \]
5. Define the relativistic kinetic energy $K$ of the particle to be $K = W$, and define the total energy $E$ to be
  
  \[ E = \dfrac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}} ~. \]
  
  So by part (d), $K = E - mc^2$. Show that
  
  \[E^2 ~=~ p^2 c^2 ~+~ (mc^2)^2 ~.\]
  
  (Hint: Expand the right side of that equation.)
6. What is $E$ when the particle is at rest?
A median of a triangle is a line segment from a vertex to the midpoint of the opposite side, and the three medians intersect at a common point. Show that this point is a triangle’s center of gravity.

For a proof that such a point exists, see p.206 in Brown, F.L., Engineering Mechanics, 2nd ed., New York: John Wiley & Sons, Inc., 1942. Some texts use the terms “center of mass” or “centroid” instead of “center of gravity,” and there are differences in the meanings. However, for the situation presented here, where the gravitational field is assumed to have constant magnitude and direction throughout the region, they all mean the same thing. ↩
Note how this is different than claiming that $F$ is “essentially constant over small intervals,” as most textbooks do. Instead, the additional infinitesimal force beyond $F(x)$ contributes zero work over $\ival{x}{x+\dx}$. ↩