다중적분의 응용(질량·질량중심)

(a) The side of the triangle from $(-a,0)$ to $(0,c)$ is straight line that passes through those two points. As $y=0$ when $x=-a$, the line must have an equation of the form $y=K(x+a)$ for some constant $K$. Since $y=c$ when $x=0$, the constant $K=\frac{c}{a}$. So that the equation is $y=\frac{c}{a}(x+a)$. has equation $cx-ay=-ac$. Similarly the side of the triangle from $(b,0)$ to $(0,c)$ has equation $y=\frac{c}{b}(b-x)$. The triangle has area $A=\frac{1}{2}(a+b)c$. It has centre of mass $(\bar x,\bar y)$ with

$$\begin{aligned} \bar x=\frac{1}{A}\iint _T x\ \mathrm{d}x\mathrm{d}y\qquad \bar y=\frac{1}{A}\iint _T y\ \mathrm{d}x\mathrm{d}y \end{aligned}$$

To evaluate the integrals we'll decompose the triangle into vertical strips as in the figure

$$\begin{aligned} \bar x&=\frac{1}{A}\iint _T x\ \mathrm{d}x\mathrm{d}y \\ &=\frac{1}{A}\bigg(\int_{-a}^0 \mathrm{d}x\int_0^{c+{c\over a}x}\mathrm{d}y\ x +\int_0^b \mathrm{d}x\int_0^{c-{c\over b}x}\mathrm{d}y\ x\bigg)\\ &=\frac{1}{A}\bigg(\int_{-a}^0 \mathrm{d}x\ x\left(c+\frac{c}{a}x\right) +\int_0^b \mathrm{d}x\ x\left(c-\frac{c}{b}x\right)\bigg)\\ &=\frac{1}{A}\left(\left[\frac{1}{2} cx^2+\frac{c}{3a}x^3\right]_{-a}^0 +\left[\frac{1}{2} cx^2-\frac{c}{3b}x^3\right]_0^b\right) \\ &=2\frac{\frac{1}{2} c(b^2-a^2)+\frac{c}{3}(a^2-b^2)}{(a+b)c} =\frac{1}{3}(b-a)\\ \bar y&=\frac{1}{A}\iint _T y\ \mathrm{d}x\mathrm{d}y \\ &=\frac{1}{A}\bigg(\int_{-a}^0 \mathrm{d}x\int_0^{c+{c\over a}x}\mathrm{d}y\ y +\int_0^b \mathrm{d}x\int_0^{c-{c\over b}x}\mathrm{d}y\ y\bigg)\\ &=\frac{1}{A}\bigg(\int_{-a}^0 \mathrm{d}x\ \frac{1}{2}\left(c+\frac{c}{a}x\right)^2 +\int_0^b \mathrm{d}x\ \frac{1}{2}\left(c-\frac{c}{b}x\right)^2\bigg)\\ &=\frac{1}{A}\left(\frac{a}{6c}\left[c+\frac{c}{a}x\right]^3\bigg|_{-a}^0 -\frac{b}{6c}\left(c-\frac{c}{b}x\right)^3\bigg|_0^b\, \right) \\ &=2\frac{\frac{ac^2}{6}+\frac{bc^2}{6}}{(a+b)c} =\frac{c}{3} \end{aligned}$$

(b) The midpoint of the side opposite $(-a,0)$ is $\frac{1}{2}\big[(b,0)+(0,c)\big]=\frac{1}{2}(b,c)$. The vector from $(-a,0)$ to $\frac{1}{2}(b,c)$ is $\frac{1}{2}\left< b,c\right>-\left<-a,0\right> =\left< a+\frac{b}{2},\frac{c}{2}\right>$. So the line joining these two points has vector parametric equation

$$\mathbf{r}(t)=\left< -a,0\right>+t\left< a+\frac{1}{2} b\,,\,\frac{1}{2} c\right>$$

The point $(\bar x,\bar y)$ lies on this line since

$$\mathbf{r}\left(\frac{2}{3}\right)=\left(\frac{1}{3}(b-a)\,,\,\frac{c}{3}\right) =(\bar x,\bar y)$$

Similarly, the midpoint of the side opposite $(b,0)$ is $\frac{1}{2}(-a,c)$. The line joining these two points has vector parametric equation

$$\mathbf{r}(t)=\left< b,0\right> +t\left<-b-\frac{1}{2} a\,,\,\frac{1}{2} c\right>$$

The point $(\bar x,\bar y)$ lies on this line too, since

$$\mathbf{r}\left(\frac{2}{3}\right)=\left(\frac{1}{3}(b-a),\frac{c}{3}\right) =(\bar x,\bar y)$$

It is not really necessary to check that $(\bar x,\bar y)$ lies on the third median, but let's do it anyway. The midpoint of the side opposite $(0,c)$ is $\frac{1}{2}(b-a,0)$. The line joining these two points has vector parametric equation

$$\mathbf{r}(t)=\left< 0,c\right>+t\left<\frac{b}{2}-\frac{a}{2},-c\right>$$

The point $(\bar x,\bar y)$ lies on this median too, since

$$\mathbf{r}\left(\frac{2}{3}\right)=\left(\frac{1}{3}(b-a),\frac{c}{3}\right) =(\bar x,\bar y)$$