Center of Mass — Question 2

Question 2

A lamina lies between $y=x$ and $y=x^2$, $0\le x\le1$, with density $\rho(x)=1+x$. Find its mass and $\bar x$.

Question 2 – Solution

Step 1: Identify the strip dimensions. On $0\le x\le1$, the upper curve is $y=x$ and the lower curve is $y=x^2$. Thus $$dA=(x-x^2)\,dx,\qquad dm=\rho(x)\,dA=(1+x)(x-x^2)\,dx.$$

Step 2: Find the mass. Since $(1+x)(x-x^2)=x-x^3$, $$\begin{align*} m&=\int_0^1(x-x^3)\,dx =\left[\frac{x^2}{2}-\frac{x^4}{4}\right]_0^1 =\frac14. \end{align*}$$

Step 3: Find the moment about the $y$-axis. The moment arm is $x$, so $$\begin{align*} M_y&=\int_0^1x\,dm =\int_0^1x(x-x^3)\,dx\\ &=\int_0^1(x^2-x^4)\,dx =\left[\frac{x^3}{3}-\frac{x^5}{5}\right]_0^1 =\frac{2}{15}. \end{align*}$$

Step 4: Compute the $x$-coordinate of the center of mass. $$\bar x=\frac{M_y}{m} =\frac{\frac{2}{15}}{\frac14} =\frac{8}{15}.$$ $$\boxed{m=\frac14,\qquad \bar x=\frac{8}{15}}$$