Science:Math Exam Resources/Courses/MATH101/April 2015/Question 08 (b)/Solution 1

Since the region is symmetric around the y-axis so we conclude that ${\bar {x}}=0$.

(Indeed, $a=-b$ and $f$ is even function, so that from the formula for ${\bar {x}}$ we can easily see this.)

To find the y-coordinate ${\bar {y}}$ we use the formula given in Hint (1). Note that from part (a) the area of the region is $A={\frac {9\pi }{4}}$ and $a=-{\frac {3}{2}},b={\frac {3}{2}}$

${\begin{aligned}{\bar {y}}={\frac {1}{2A}}\int _{a}^{b}(f(x))^{2}\,dx&={\frac {1}{2}}\cdot {\frac {4}{9\pi }}\int _{-{\frac {3}{2}}}^{\frac {3}{2}}9-4x^{2}\,dx&={\frac {1}{2}}\cdot {\frac {4}{9\pi }}\cdot 2\int _{0}^{\frac {3}{2}}9-4x^{2}\,dx\\&={\frac {4}{9\pi }}\left(9x-{\frac {4}{3}}x^{3}\right){\bigg |}_{0}^{\frac {3}{2}}\\&={\frac {4}{9\pi }}\left(9\cdot {\frac {3}{2}}-{\frac {4}{3}}\left({\frac {3}{2}}\right)^{3}\right)\\&={\frac {4}{9\pi }}\left({\frac {27}{2}}-{\frac {9}{2}}\right)\\&={\frac {4}{\pi }}\end{aligned}}$

Center of mass: ${\color {blue}\left(0,{\frac {4}{\pi }}\right)}$