다중적분의 응용(질량·질량중심)

(a) $\iint _D x\ \mathrm{d}x\,\mathrm{d}y=0$ because $x$ is odd under $x\rightarrow-x$, i.e. under reflection about the $y$–axis, while the domain of integration is symmetric about the $y$–axis. $\iint _D 3\ \mathrm{d}x\,\mathrm{d}y$ is the three times the area of a half disc of radius $2$. So, $\iint _D(x+3)\mathrm{d}x\,\mathrm{d}y=3\times \tfrac{1}{2}\times\pi 2^2=6\pi$.

(s) $\iint _R x\ \mathrm{d}x\,\mathrm{d}y/\iint _R \mathrm{d}x\,\mathrm{d}y$ is the average value of $x$ in the rectangle $R$, namely $\frac{a}{2}$. Similarly, $\iint _R y\ \mathrm{d}x\,\mathrm{d}y/\iint _R \mathrm{d}x\,\mathrm{d}y$ is the average value of $y$ in the rectangle $R$, namely $\frac{b}{2}$. $\iint _R \mathrm{d}x\,\mathrm{d}y$ is area of the rectangle $R$, namely $ab$. So,

• $\iint _R x\ \mathrm{d}x\,\mathrm{d}y =\frac{a}{2} \iint _R \mathrm{d}x\,\mathrm{d}y =\frac{a}{2} ab$ and
• $\iint _R y\ \mathrm{d}x\,\mathrm{d}y =\frac{b}{2} \iint _R \mathrm{d}x\,\mathrm{d}y =\frac{b}{2} ab$

and $\iint _R (x+y)\mathrm{d}x\,\mathrm{d}y=\tfrac{1}{2} ab (a+b)$.

Here is a sketch of $D$.

By definition, the centre of mass is $(\bar x, \bar y)$, with $\bar x$ and $\bar y$ being the weighted averages of the $x$ and $y$–coordinates, respectively, over $D$. That is,

$$\begin{aligned} \bar x = \frac{\iint _D x\ \rho(x,y)\ \mathrm{d}A}{\iint _D \rho(x,y)\ \mathrm{d}A} \qquad \bar y = \frac{\iint _D y\ \rho(x,y)\ \mathrm{d}A}{\iint _D \rho(x,y)\ \mathrm{d}A} \end{aligned}$$

By symmetry under reflection in the $y$–axis, we have $\bar x=0$. So we just have to determine $\bar y$. We'll evaluate the integrals using vertical strips as in the figure above. Looking at that figure, we see that

• $x$ runs from $-1$ to $1$, and
• for each fixed $x$ in that range, $y$ runs from $x^2$ to $1$.

So the denominator is

$$\begin{aligned} \iint _D \rho(x,y)\ \mathrm{d}A &= \int_{-1}^1\mathrm{d}x\int_{x^2}^1\mathrm{d}y\ \overbrace{y}^{\rho(x,y)} \\ &= \frac{1}{2}\int_{-1}^1 \mathrm{d}x\ (1-x^4) =\int_0^1 \mathrm{d}x\ (1-x^4) \\ &=\frac{4}{5} \end{aligned}$$

and the numerator of $\bar y$ is

$$\begin{aligned} \iint _D y\,\rho(x,y)\ \mathrm{d}A &= \int_{-1}^1\mathrm{d}x\int_{x^2}^1\mathrm{d}y\ y\overbrace{y}^{\rho(x,y)} \\ &= \frac{1}{3}\int_{-1}^1 \mathrm{d}x\ (1-x^6) =\frac{2}{3}\int_0^1 \mathrm{d}x\ (1-x^6) \\ &=\frac{2}{3}\ \frac{6}{7} = \frac{4}{7} \end{aligned}$$

All together, $\bar x=0$ and

$$\begin{aligned} \bar y = \frac{\frac{4}{7}} {\frac{4}{5}} = \frac{5}{7} \end{aligned}$$

(a) Here is a sketch of $R$.

(b) Considering that

• $\rho(x,y)$ is invariant under rotations about the origin and
• the outer curve $x^2+y^2=4$ is invariant under rotations about the origin and
• the given hint involves a $\theta$ integral

we'll use polar coordinates.

Observe that the line $x=1$ and the circle $x^2+y^2=4$ intersect when

$$1+y^2=4 \iff y=\pm\sqrt{3}$$

and that the polar coordinates of the point $(x,y)=\big(1,\sqrt{3}\big)$ are $r=\sqrt{x^2+y^2}=2$ and $\theta=\arctan\frac{y}{x}=\arctan \sqrt{3} =\frac{\pi}{3}$. Looking at the sketch

we see that, on $R$,

• $\theta$ runs from $-\frac{\pi}{3}$ to $\frac{\pi}{3}$ and
• for each fixed $\theta$ in that range, $r$ runs from $\frac{1}{\cos\theta} =\sec\theta$ to $2$.
• In polar coordinates, $\mathrm{d}A=r\,\mathrm{d}r\,\mathrm{d}\theta$, and
• the density $\rho =\frac{1}{\sqrt{x^2+y^2}} =\frac{1}{r}$

So the mass is

$$\begin{aligned} M&=\iint _R \rho(x,y)\ \mathrm{d}A =\int_{-\pi/3}^{\pi/3}\mathrm{d}\theta\int_{\sec\theta}^2\mathrm{d}r\ \frac{r}{r} =\int_{-\pi/3}^{\pi/3}\mathrm{d}\theta\ \big[2-\sec\theta\big] \\ &=2\int_0^{\pi/3}\mathrm{d}\theta\ \big[2-\sec\theta\big] \\ &= 2\Big[2\theta - \ln\big(\sec\theta+\tan\theta\big)\Big]_0^{\pi/3} \\ &= 2\left[\frac{2\pi}{3} - \ln\big(2+\sqrt{3}\big) + \ln\big(1+0\big)\right] \\ &= \frac{4\pi}{3} - 2\ln\big(2+\sqrt{3}\big) \end{aligned}$$

(c) By definition, the centre of mass is $(\bar x, \bar y)$, with $\bar x$ and $\bar y$ being the weighted averages of the $x$ and $y$–coordinates, respectively, over $R$. That is,

$$\begin{aligned} \bar x = \frac{\iint _R x\ \rho(x,y)\ \mathrm{d}A}{\iint _R \rho(x,y)\ \mathrm{d}A} \qquad \bar y = \frac{\iint _R y\ \rho(x,y)\ \mathrm{d}A}{\iint _R \rho(x,y)\ \mathrm{d}A} \end{aligned}$$

By symmetry under reflection in the $x$–axis, we have $\bar y=0$. So we just have to determine $\bar x$. The numerator is

$$\begin{aligned} \iint _R x\ \rho(x,y)\ \mathrm{d}A &=\int_{-\pi/3}^{\pi/3}\mathrm{d}\theta\int_{\sec\theta}^2\mathrm{d}r\ \frac{r}{r}\ \overbrace{r\cos\theta}^{x} \\ &=\frac{1}{2}\int_{-\pi/3}^{\pi/3}\mathrm{d}\theta\ \big[4-\sec^2\theta\big] \cos\theta =\int_0^{\pi/3}\mathrm{d}\theta\ \big[4\cos\theta-\sec\theta\big] \\ &= \Big[4\sin\theta - \ln\big(\sec\theta+\tan\theta\big)\Big]_0^{\pi/3} \\ &= \left[4\frac{\sqrt{3}}{2} - \ln\big(2+\sqrt{3}\big) + \ln\big(1+0\big)\right] \\ &= 2\sqrt{3}- \ln\big(2+\sqrt{3}\big) \end{aligned}$$

All together, $\bar y=0$ and

$$\begin{aligned} \bar x = \frac{2\sqrt{3}- \ln\big(2+\sqrt{3}\big)} {\frac{4\pi}{3} - 2\ln\big(2+\sqrt{3}\big)} \approx 1.38 \end{aligned}$$

Call the plate $P$. By definition, the centre of mass is $(\bar x, \bar y)$, with $\bar x$ and $\bar y$ being the weighted averages of the $x$ and $y$–coordinates, respectively, over $P$. That is,

$$\begin{aligned} \bar x = \frac{\iint _P x\ \rho(x,y)\ \mathrm{d}A}{\iint _P \rho(x,y)\ \mathrm{d}A} \qquad \bar y = \frac{\iint _P y\ \rho(x,y)\ \mathrm{d}A}{\iint _P \rho(x,y)\ \mathrm{d}A} \end{aligned}$$

with $\rho(x,y)=k$. Here is a sketch of $P$.

By symmetry under reflection in the line $y=x$, we have $\bar y=\bar x$. So we just have to determine

$$\bar x = \frac{\iint _P x\ \mathrm{d}A}{\iint _P \mathrm{d}A}$$

The denominator is just one quarter of the area of circular disk of radius $1$. That is, $\iint _P \mathrm{d}A=\frac{\pi}{4}$. We'll evaluate the numerator using polar coordinates as in the figure above. Looking at that figure, we see that

• $\theta$ runs from $0$ to $\frac{\pi}{2}$, and
• for each fixed $\theta$ in that range, $r$ runs from $0$ to $1$.

As $\mathrm{d}A=r\,\mathrm{d}r\,\mathrm{d}\theta$, and $x=r\cos\theta$, the numerator

$$\begin{aligned} \iint _P x\ \mathrm{d}A &=\int_0^{\pi/2}\mathrm{d}\theta \int_0^1\mathrm{d}r\ r\overbrace{r\cos\theta}^{x} =\left[\int_0^{\pi/2}\mathrm{d}\theta\ \cos\theta\right] \left[ \int_0^1\mathrm{d}r\ r^2\right] \\ &=\Big[\sin\theta\Big]_0^{\pi/2} \left[\frac{r^3}{3}\right]_0^1 \\ &=\frac{1}{3} \end{aligned}$$

All together

$$\begin{aligned} \bar x = \bar y = \frac{1/3}{\pi/4} =\frac{4}{3\pi} \end{aligned}$$

Here is a sketch of $R$.

Note that

• the equation of the straight line through $(2,0)$ and $(0,2)$ is $y=2-x$, or $x=2-y$. (As a check note that both points $(2,0)$ and $(0,2)$ are on $x=2-y$.
• The equation of the straight line through $(1,0)$ and $(0,2)$ is $y=2-2x$, or $x=\frac{2-y}{2}$. (As a check note that both points $(0,2)$ and $(1,0)$ are on $x=\frac{2-y}{2}$.

By definition, the $y$–coordinate of the center of mass of $R$ is the weighted average of $y$ over $R$, which is

$$\bar y =\frac{\iint _R y\,\rho(x,y)\,\mathrm{d}A}{\iint _R \rho(x,y)\,\mathrm{d}A} =\frac{\iint _R y^3\,\mathrm{d}A}{\iint _R y^2\,\mathrm{d}A}$$

On $R$,

• $y$ runs from $0$ to $2$. That is, $0\le y\le 2$.
• For each fixed $y$ in that range, $x$ runs from $\frac{2-y}{2}$ to $2-y$. In inequalities, that is $\frac{2-y}{2}\le x\le 2-y$.

Thus

$$R = \left\{\ (x,y)\ \left|\ 0\le y\le 2,\ \frac{2-y}{2}\le x\le 2-y \right.\right\}$$

For both $n=2$ and $n=3$, we have

$$\begin{aligned} \iint _R y^n\,\mathrm{d}A &=\int_0^2\mathrm{d}y \int_{\frac{2-y}{2}}^{2-y}\mathrm{d}x\ y^n \\ &=\int_0^2\mathrm{d}y \ y^n\frac{2-y}{2} \\ &=\frac{1}{2}\left[\frac{2y^{n+1}}{n+1}-\frac{y^{n+2}}{n+2}\right]_0^2 \\ &=\frac{1}{2}\left[\frac{2^{n+2}}{n+1}-\frac{2^{n+2}}{n+2}\right] \\ &=\frac{2^{n+1}}{(n+1)(n+2)} \end{aligned}$$

So

$$\begin{aligned} \bar y =\frac{\iint _R y^3\,\mathrm{d}A}{\iint _R y^2\,\mathrm{d}A} =\frac{ \frac{2^4}{(4)(5)} }{ \frac{2^{3}}{(3)(4)} } =\frac{6}{5} \end{aligned}$$

By the definition given in the statement with $(a,b)=(0,0)$, the average is

$$\begin{aligned} \frac{1}{A(D)}\iint _D \sqrt{x^2+y^2}\ \mathrm{d}x\,\mathrm{d}y \end{aligned}$$

The denominator $A(D) = \pi$. We'll use polar coordinates to evaluate the numerator.

$$\begin{aligned} \iint _D \sqrt{x^2+y^2}\ \mathrm{d}x\,\mathrm{d}y &=\int_0^{2\pi}\mathrm{d}\theta\int_0^1\mathrm{d}r \ r\sqrt{r^2\cos^2\theta+r^2\sin^2\theta} \\ &=\int_0^{2\pi}\mathrm{d}\theta\int_0^1\mathrm{d}r\ r^2 =\int_0^{2\pi}\mathrm{d}\theta\ \frac{1}{3} \\ &=\frac{2\pi}{3} \end{aligned}$$

So the average is

$$\frac{\frac{2\pi}{3}}{\pi}=\frac{2}{3}$$