# 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 ${\displaystyle {\bar {x}}=0}$.
(Indeed, ${\displaystyle a=-b}$ and ${\displaystyle f}$ is even function, so that from the formula for ${\displaystyle {\bar {x}}}$ we can easily see this.)
To find the y-coordinate ${\displaystyle {\bar {y}}}$ we use the formula given in Hint (1). Note that from part (a) the area of the region is ${\displaystyle A={\frac {9\pi }{4}}}$ and ${\displaystyle a=-{\frac {3}{2}},b={\frac {3}{2}}}$
{\displaystyle {\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: ${\displaystyle {\color {blue}\left(0,{\frac {4}{\pi }}\right)}}$