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

$\begin{array}{rl}\underset{0}{\overset{2}{\int }}x((6-x)-x^2)dx& =\underset{0}{\overset{2}{\int }}(6x-x^2-x^3)dx\\ & =(3x^2-\frac{x^3}{3}-\frac{x^4}{4}){|}_0^2\\ & =12-\frac{8}{3}-4=\frac{16}{3},\end{array}$

while the vertical moment is

$\begin{array}{rl}\frac{1}{2}\underset{0}{\overset{2}{\int }}((6-x{)}^2-x^4)dx& =\frac{1}{2}(-\frac{(6-x{)}^3}{3}-\frac{x^5}{5}){|}_0^2\\ & =\frac{1}{2}(\frac{216-64}{3}-\frac{32}{5})=\frac{332}{15}.\end{array}$

Note that the last integral could also be evaluated as

$\begin{array}{rl}\frac{1}{2}\underset{0}{\overset{2}{\int }}(36-12x+x^2-x^4)dx& =\frac{1}{2}(36x-6x^2+\frac{x^3}{3}-\frac{x^5}{5}){|}_0^2\\ & =\frac{1}{2}(72-24+\frac{8}{3}-\frac{32}{5})=\frac{332}{15}.\end{array}$

Using the fact that

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

the centroid is therefore

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