Center of Mass — Question 1

Question 1 – Solution

Step 1: Write the upper boundary as a function of $x$. From $x+2y=6$, $$y=f(x)=3-\frac{x}{2},\qquad 0\le x\le6.$$

Step 2: Find the area (and hence the mass for unit density). $$\begin{align*} A&=\int_0^6 f(x)\,dx =\int_0^6\left(3-\frac{x}{2}\right)dx\\ &=\left[3x-\frac{x^2}{4}\right]_0^6=18-9=9. \end{align*}$$

Step 3: Find the moment about the $y$-axis. $$\begin{align*} M_y&=\int_0^6 x f(x)\,dx =\int_0^6\left(3x-\frac{x^2}{2}\right)dx\\ &=\left[\frac{3x^2}{2}-\frac{x^3}{6}\right]_0^6=54-36=18. \end{align*}$$

Step 4: Find the moment about the $x$-axis. Each vertical strip has centroid height $f(x)/2$, so $$\begin{align*} M_x&=\frac12\int_0^6 f(x)^2\,dx =\frac12\int_0^6\left(3-\frac{x}{2}\right)^2dx=9. \end{align*}$$

Step 5: Divide each moment by the area. $$\bar x=\frac{M_y}{A}=\frac{18}{9}=2,\qquad \bar y=\frac{M_x}{A}=\frac{9}{9}=1.$$ $$\boxed{(\bar x,\bar y)=(2,1)}$$