§15.4 이중적분의 응용Applications of Double Integrals

Solution. The triangular region $D$ lies between the lower boundary $y=1-x$ and the upper boundary $y=1$ as $x$ runs from $0$ to $1$, so as a type I region,

Substituting $\sigma(x,y)=xy$ into Equation (2) gives

Treating $x$ as constant in the inner integral,

using $1-(1-x)^2=2x-x^2$. Integrating this over $x$,

Hence the total charge on $D$ is $Q=\dfrac{5}{24}$ C.

Solution. Of the three vertices $(0,0)$, $(1,0)$, $(0,2)$, the hypotenuse joins $(1,0)$ and $(0,2)$ and has equation $\dfrac{x}{1}+\dfrac{y}{2}=1$, i.e. $y=2-2x$. So the triangle is the type I region

The mass is

Writing $u=2-2x$, the inner integral is

(expanding the two terms, the linear terms in $x$ cancel). So

The moment $M_y$ comes from multiplying the same inner-integral result by $x$:

The moment $M_x$ requires recomputing the inner integral with an extra factor $y$:

and integrating over $x$ from $0$ to $1$,

Since $\bar x=M_y/m$ and $\bar y=M_x/m$,

Hence the mass is $m=\dfrac83$, and the center of mass is $(\bar x,\bar y)=\left(\dfrac38,\dfrac{11}{16}\right)$.

Solution. Place the center of the circle at the origin. Since the density is proportional to the distance from the origin, $\rho(x,y)=K\sqrt{x^2+y^2}$ for a proportionality constant $K$, and the region is the upper half-disk

Because the density depends only on the distance from the origin and $D$ is bounded by an arc centered there, it is natural to switch to polar coordinates, $D=\{(r,\theta)\mid 0\le r\le a,\ 0\le\theta\le\pi\}$. With $\rho=Kr$ and $dA=r\,dr\,d\theta$, the mass is

Since $D$ is symmetric about the $y$-axis and the density $Kr$ is unchanged under $x\mapsto-x$, $M_y=0$, so $\bar x=0$. For $M_x$,

The center of mass is therefore

so the mass is $m=\dfrac{K\pi a^3}{3}$ and the center of mass is $(\bar x,\bar y)=\left(0,\dfrac{3a}{2\pi}\right)$. This lies farther from the origin than the center of mass $\left(0,\dfrac{4a}{3\pi}\right)$ of the uniform-density case, since density growing with $r$ shifts more of the mass toward the outer part of the disk.

Solution. Here $D=\{(x,y)\mid x^2+y^2\le a^2\}$ and the density $\rho$ is constant. Compute the polar moment of inertia $I_0$ first and recover $I_x,I_y$ from symmetry. In polar coordinates $D=\{(r,\theta)\mid 0\le r\le a,\ 0\le\theta\le2\pi\}$ and $x^2+y^2=r^2$, so

Reflecting the disk in the line $y=x$ — that is, swapping $x$ and $y$ — leaves both $D$ and the constant density $\rho$ unchanged, so the integrands $x^2\rho$ and $y^2\rho$ give the same integral over $D$: $I_x=I_y$. Substituting into Equation (8), $I_0=I_x+I_y$, gives

Hence $I_x=I_y=\dfrac{\rho\pi a^4}{4}$ and $I_0=\dfrac{\rho\pi a^4}{2}$.

Solution. The disk of Example 4 has area $\pi a^2$ and constant density $\rho$, so its mass is $m=\rho\pi a^2$. Its moment of inertia about the $x$-axis was found to be $I_x=\dfrac{\rho\pi a^4}{4}$. Substituting these into Equation (10), $m\bar{\bar y}^{\,2}=I_x$, gives

Taking the positive square root, since $\bar{\bar y}\gt0$,

Hence the radius of gyration about the $x$-axis is $\bar{\bar y}=\dfrac{a}{2}$ — concentrating the entire mass of the disk at half its radius from the $x$-axis reproduces the same $I_x$.

Solution. For $f$ to be a joint density function, its integral over the whole plane must equal $1$; since $f$ vanishes outside $[0,10]\times[0,10]$,

The inner integral is

and integrating over $x$,

So $1500C=1$, giving $C=\dfrac{1}{1500}$.

For the probability, intersecting $x\le7$, $y\ge2$ with $[0,10]\times[0,10]$ leaves the rectangle $0\le x\le7$, $2\le y\le10$, so

The inner integral is

and

so

Hence $C=\dfrac{1}{1500}$ and $P(X\le7,\,Y\ge2)=\dfrac{868}{1500}\approx0.5787$.

Solution. Since $X$ and $Y$ are independent and exponentially distributed with means $10$ and $5$ minutes, their individual densities are $f_1(x)=\tfrac1{10}e^{-x/10}$ ($x\ge0$) and $f_2(y)=\tfrac15e^{-y/5}$ ($y\ge0$), so the joint density is their product

(and $0$ elsewhere). The probability sought is the integral of $f$ over $D=\{(x,y)\mid x\ge0,\ y\ge0,\ x+y\lt20\}$, the triangular region with legs of length $20$. Writing $D$ as the type I region $0\le x\le20$, $0\le y\le20-x$,

The inner integral is

so writing $e^{-(20-x)/5}=e^{-4}e^{x/5}$,

The first term is $\Big[-e^{-x/10}\Big]_0^{20}=1-e^{-2}$, and the second integral is $\Big[e^{x/10}\Big]_0^{20}=e^2-1$, so

Numerically $e^{-2}\approx0.1353$ and $e^{-4}\approx0.0183$, so $P(X+Y\lt20)\approx1-0.2707+0.0183=0.7476$. Hence $P(X+Y\lt20)=1+e^{-4}-2e^{-2}\approx0.7476$, about $75\%$.

Solution. (a) Since $X$ and $Y$ are independent and normally distributed with means $4.0$ and $6.0$ and the same standard deviation $\sigma=0.01$, the joint density is the product of the individual densities,

With $\sigma=0.01$, $2\sigma^2=0.0002=1/5000$ and $2\pi\sigma^2=2\pi\cdot0.0001$, so $1/(2\pi\sigma^2)=5000/\pi$ and

(b) The complement of “the diameter or the length deviates from its mean by more than $0.02$ cm” is “both the diameter and the length stay within $0.02$ cm of their means,” so the desired probability is $1-P(3.98\lt X\lt4.02,\ 5.98\lt Y\lt6.02)$. Standardizing $X$ gives a standard normal $Z_1=(X-4)/0.01$, and since $0.02=2\sigma$,

where $\Phi$ is the standard normal cumulative distribution function. $Y$ has the same standard deviation, so it gives the same probability $2\Phi(2)-1$, and independence turns the joint event into a product:

From tables $\Phi(2)\approx0.97725$, so $2\Phi(2)-1\approx0.95450$ and its square is $\approx0.91107$. Hence

Rounded to two decimals this is $\approx0.09$ — the same value obtained by rounding the joint probability to $0.91$ first, as a calculator would, and then taking the complement.