Science:Math Exam Resources/Courses/MATH101/April 2011/Question 03 (b)/Solution 1

First we compute the area A of the shape. This is given by

$\displaystyle {\begin{aligned}A&=\int _{-\pi /2}^{\pi /2}1+\sin(x)dx\\&=x{\big |}_{-\pi /2}^{\pi /2}-\cos {\big |}_{-\pi /2}^{\pi /2}\\&=\pi -\cos(\pi /2)+\cos(-\pi /2)\\&=\pi \end{aligned}}$

The coordinates of the centroid are computed using the formulas

$\displaystyle {{\bar {x}}={\frac {1}{\pi }}\int _{-\pi /2}^{\pi /2}x(1+\sin(x))dx}$

and

$\displaystyle {{\bar {y}}={\frac {1}{\pi }}\int _{-\pi /2}^{\pi /2}{\frac {1}{2}}(1+\sin(x))^{2}dx}$

We will compute ${\bar {x}}$ first.

$\displaystyle {\begin{aligned}{\bar {x}}&={\frac {1}{\pi }}\int _{-\pi /2}^{\pi /2}x+x\sin(x)dx\\&={\frac {1}{\pi }}\int _{-\pi /2}^{\pi /2}xdx+{\frac {1}{\pi }}\int _{-\pi /2}^{\pi /2}x\sin(x)dx\\&=0+{\frac {1}{\pi }}\int _{-\pi /2}^{\pi /2}x\sin(x)dx\end{aligned}}$

The first integral is zero because the function is odd. For the second we can use integration by parts.

$\displaystyle {\begin{aligned}\int x\sin(x)dx&=-x\cos(x)+\int \cos(x)dx\\&=-x\cos(x)+\sin(x)+C\end{aligned}}$

That means if we plug in the limits for the definite integral we get

$\displaystyle {\begin{aligned}{\bar {x}}&=1/\pi (-x\cos(x)+\sin(x)){\big |}_{-\pi /2}^{\pi /2}\\&=1/\pi (\sin(\pi /2)-\sin(-\pi /2))\\&=2/\pi \end{aligned}}$

Now we compute ${\bar {y}}$

$\displaystyle {\begin{aligned}{\bar {y}}&={\frac {1}{\pi }}\int _{-\pi /2}^{\pi /2}{\frac {1}{2}}(1+\sin(x))^{2}dx\\&={\frac {1}{2\pi }}\int _{-\pi /2}^{\pi /2}1+2\sin(x)+\sin(x)^{2}dx\\&={\frac {1}{2\pi }}{\Big (}\int _{-\pi /2}^{\pi /2}1dx+2\int _{-\pi /2}^{\pi /2}\sin(x)dx+\int _{-\pi /2}^{\pi /2}\sin(x)^{2}dx{\Big )}\\&={\frac {1}{2}}+{\frac {1}{2\pi }}\int _{-\pi /2}^{\pi /2}\sin(x)^{2}dx\end{aligned}}$

To compute this we note $\sin(x)^{2}={\frac {1-\cos(2x)}{2}}$

$\displaystyle {\begin{aligned}\int \sin(x)^{2}dx&=\int {\frac {1-\cos(2x)}{2}}dx\\&={\frac {x}{2}}-{\frac {1}{4}}\sin(2x)+C\end{aligned}}$

Hence

$\displaystyle {\begin{aligned}{\bar {y}}&={\frac {1}{2}}+{\frac {1}{2\pi }}({\frac {x}{2}}-{\frac {1}{4}}\sin(2x))_{-\pi /2}^{\pi /2}\\&={\frac {3}{4}}\end{aligned}}$

Thus the centroid is $({\frac {2}{\pi }},{\frac {3}{4}})$