Center of Mass — Question 10

Question 10

A symmetric lamina has density $\rho(x,y)=1+y$ on $-1\le x\le1, 0\le y\le1-x^2$. State $\bar x$, then set up exact integrals for $m$ and $\bar y$.

Question 10 – Solution

Step 1: Use symmetry. The region is symmetric about the $y$-axis, and $\rho(x,y)=1+y$ is unchanged when $x$ is replaced by $-x$. Therefore, $$\boxed{\bar x=0}.$$

Step 2: Set up and evaluate the mass. Let $h=1-x^2$. Then $$\begin{align*} m&=\int_{-1}^1\int_0^{1-x^2}(1+y)\,dy\,dx\\ &=\int_{-1}^1\left[h+\frac{h^2}{2}\right]dx\\ &=\int_{-1}^1\left[(1-x^2)+\frac12(1-x^2)^2\right]dx =\frac{28}{15}. \end{align*}$$

Step 3: Set up and evaluate the moment about the $x$-axis. $$\begin{align*} M_x&=\int_{-1}^1\int_0^{1-x^2}y(1+y)\,dy\,dx\\ &=\int_{-1}^1\left[\frac{h^2}{2}+\frac{h^3}{3}\right]dx\\ &=\frac12\int_{-1}^1(1-x^2)^2dx +\frac13\int_{-1}^1(1-x^2)^3dx\\ &=\frac{8}{15}+\frac{32}{105}=\frac{88}{105}. \end{align*}$$

Step 4: Compute $\bar y$. $$\bar y=\frac{M_x}{m} =\frac{\frac{88}{105}}{\frac{28}{15}} =\frac{22}{49}.$$ $$\boxed{(\bar x,\bar y)=\left(0,\frac{22}{49}\right)}$$