Center of Mass — Question 7

Question 7 – Solution

Step 1: Use symmetry before integrating. Both the region and its uniform density are symmetric about the $y$-axis. Therefore, $$\bar x=0.$$

Step 2: Find the area. $$\begin{align*} A&=\int_{-2}^{2}(4-x^2)\,dx =2\int_0^2(4-x^2)\,dx\\ &=2\left[4x-\frac{x^3}{3}\right]_0^2 =2\left(8-\frac83\right)=\frac{32}{3}. \end{align*}$$

Step 3: Find the moment about the $x$-axis. For a vertical strip under $f(x)=4-x^2$, $$\begin{align*} M_x&=\frac12\int_{-2}^{2}f(x)^2\,dx =\int_0^2(4-x^2)^2\,dx\\ &=\int_0^2(16-8x^2+x^4)\,dx\\ &=\left[16x-\frac{8x^3}{3}+\frac{x^5}{5}\right]_0^2 =\frac{256}{15}. \end{align*}$$

Step 4: Compute $\bar y$. $$\bar y=\frac{M_x}{A} =\frac{\frac{256}{15}}{\frac{32}{3}} =\frac85.$$ $$\boxed{(\bar x,\bar y)=\left(0,\frac85\right)}$$