Science:Math Exam Resources/Courses/MATH101/April 2016/Question 08 (a)/Solution 1

Setting $x^{2}=6-x$, we see that the two curves intersect at $x=2$ (as well as $x=-3$, which is to the left of the $y$-axis and therefore irrelevant).

The horizontal moment is

${\begin{aligned}\int _{0}^{2}x{\big (}(6-x)-x^{2}{\big )}\,dx&=\int _{0}^{2}(6x-x^{2}-x^{3})\,dx\\&={\bigg (}3x^{2}-{\frac {x^{3}}{3}}-{\frac {x^{4}}{4}}{\bigg )}{\bigg |}_{0}^{2}\\&=12-{\frac {8}{3}}-4={\frac {16}{3}},\end{aligned}}$

while the vertical moment is

${\begin{aligned}{\frac {1}{2}}\int _{0}^{2}{\big (}(6-x)^{2}-x^{4}\ {\big )}\,dx&={\frac {1}{2}}{\bigg (}{-}{\frac {(6-x)^{3}}{3}}-{\frac {x^{5}}{5}}{\bigg )}{\bigg |}_{0}^{2}\\&={\frac {1}{2}}{\bigg (}{\frac {216-64}{3}}-{\frac {32}{5}}{\bigg )}={\frac {332}{15}}.\end{aligned}}$

Note that the last integral could also be evaluated as

${\begin{aligned}{\frac {1}{2}}\int _{0}^{2}{\big (}36-12x+x^{2}-x^{4}\ {\big )}\,dx&={\frac {1}{2}}{\bigg (}36x-6x^{2}+{\frac {x^{3}}{3}}-{\frac {x^{5}}{5}}{\bigg )}{\bigg |}_{0}^{2}\\&={\frac {1}{2}}{\bigg (}72-24+{\frac {8}{3}}-{\frac {32}{5}}{\bigg )}={\frac {332}{15}}.\end{aligned}}$

Using the fact that

The centroid of area $A={\frac {1}{A}}({\text{The horizontal moment}},{\text{The vertical moment}}),$

the centroid is therefore

$\displaystyle {\frac {1}{22/3}}{\bigg (}{\frac {16}{3}},{\frac {332}{15}}{\bigg )}={\color {blue}{\bigg (}{\frac {8}{11}},{\frac {166}{55}}{\bigg )}}$.